Chapter 2 Structure of Atom of Chemistry Questions and Answers: NCERT Solutions for Class 11th Chemistry (Partnership Chemistry )

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Class 11 Chemistry (Partnership Chemistry ) Chapter 2: Structure of Atom of Chemistry - Questions and Answers of NCERT Book Solutions.

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1. (i). Calculate the Number of Electrons Which Will Together Weigh One Gram.

Ans: Mass of one electron = 9.10939×10−31kg 9.10939×10−31kg
Number of electrons that weigh 9.10939×10−31kg 9.10939×10−31kg
= 1 Number of electrons that will weigh 1 g = 1×10−3kg 1×10−3kg

19.10939×10−31kg(1×10−3kg)=0.1098×10−3+31
19.10939×10−31kg(1×10−3kg)=0.1098×10−3+31
=1.098×1027
(ii).Calculate the Mass and Charge of One Mole of Electrons.
Ans: Mass of one electron = 9.10939×10−31kg
9.10939×10−31kg
Mass of one mole of electron = (6.022×1023)×(9.10939×10−31kg) (6.022×1023)×(9.10939×10−31kg)
= 5.48×10−7kg
5.48×10−7kg Charge on one electron = 1.6022×10−19coulomb 1.6022×10−19coulomb
Charge on one mole of electron = (1.6022×10−19)(6.022×1023)
(1.6022×10−19)(6.022×1023)
= 9.65×104C

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2. (i).Calculate the Total Number of Electrons Present in One Mole of Methane.

Ans: Number of electrons present in 1 molecule of methane (CH4 CH4
)= {1(6)+4(1)}=10
{1(6)+4(1)}=10
Number of electrons present in 1 mole i.e., 6.023×1023
6.023×1023
molecules of methane = 6.023×1023×10=6.023×1024
6.023×1023×10=6.023×1024
(ii). a) Find the Total Number of Neutrons in 7mg of 14C14C . (Assume the Mass of a Neutron = 1.675×10−27kg1.675×10−27kg )
Ans: Number of atoms of 14C
14C
in 1 mole = 6.023×1023
6.023×1023
Since 1 atom of 14C
14C
contains (14-6) i.e., 8 neutrons, the number of neutrons in 14g of 14C
14CM
is (6.023×1023)×8
(6.023×1023)×8
Number of neutrons in 7 mg
=6.022×1023×8×71400mg
=6.022×1023×8×71400mg
=2.4092×1021
=2.4092×1021
(b) Find the Total Mass of Neutrons in 7mg of 14C14C . (Assume the Mass of a Neutron = 1.675×10−27kg1.675×10−27kg )
Ans: Mass of one neutron = 1.67493×10−27kg
1.67493×10−27kg
Mass of total neutrons in 7g of 14C
14C
=(2.4092×1021)(1.67493×10−27kg)
=(2.4092×1021)(1.67493×10−27kg)
=4.0352×10−6kg
=4.0352×10−6kg
(iii). (a) Find the Total Number of Protons in 34mg of NH3NH3 at STP.
Ans: 1 mole of NH3
NH3
= {1(14) +3(1)} g of NH3
NH3
= 17g of NH3
NH3
=6.023×1023
=6.023×1023
Molecules of NH3
NH3
Total number of protons present in 1 molecule of NH3
NH3
= 1(7)+1(3)1(7)+1(3) =10
Number of protons in 6.023×1023
6.023×1023
molecules of NH3
NH3

=(6.023×1023)(10)=6.023×1024
=(6.023×1023)(10)=6.023×1024
17g of NH3
NH3
contains (6.023×1024)
(6.023×1024)
protons.
Number of protons in 34mg of NH3 NH3
(b) Find the total mass of protons in 34mg of NH3NH3 at STP. Will the answer change if the temperature and pressure changed?
Ans: Mass of one proton = 1.67493×10−27kg
1.67493×10−27kg
Total mass of protons in 34mg of NH3 NH3
=(1.67493×10−27kg)(1.2046×1022)
=(1.67493×10−27kg)(1.2046×1022)
=2.0176×10−5kg
=2.0176×10−5kg
The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

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3. How Many Neutrons and Protons are There in the Following Nuclei?

Ans: 136C613C : Atomic mass =13
Atomic number = Number of Protons = 6
Number of neutrons = (Atomic Mass)-(Atomic Number)=13-6=7
168O816O :
Atomic mass =16
Atomic number = Number of Protons = 8
Number of neutrons = (Atomic Mass)-(Atomic Number) =16-8=8
2412Mg1224Mg :
Atomic mass =24
Atomic number = Number of Protons = 12
Number of neutrons = (Atomic Mass)-(Atomic Number)=24-12 =12
5626Fe2656Fe :
Atomic mass = 56
Atomic number = Number of Protons = 26
Number of neutrons = (Atomic Mass)-(Atomic Number) = 56-26=30
8838Sr3888Sr :
Atomic mass = 88
Atomic number = Number of Protons = 38
Number of neutrons = (Atomic Mass)-(Atomic Number)= 88-38=50

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4. Write the Complete Symbol for the Atom with the Given Atomic Number (Z) and Atomic mass (A)
a. Z = 17, A = 35
Ans: 3517Cl1735Cl
b. Z = 92, A = 233
Ans: 23392U92233U
c. Z = 4, A = 9

Ans: 94Be49Be

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5. Yellow light emitted from a sodium lamp has a wavelength (λλ ) of 580 nm. Calculate the frequency (υυ) and wave number ( υ−υ− ) of the yellow light.

Ans:
From the expression,
λ=cυ
λ=cυ
We get,
υ=cλυ=cλ ----- (i)
Where,
υυ = frequency of yellow light
c = velocity of light in vacuum = 3×108m/s3×108m/s
λλ = wavelength of yellow light = 580nm = 580×10−9m580×10−9m
Substituting the values in the expression (i)
υ=3×108580×10−9=5.17×1014S−1
υ=3×108580×10−9=5.17×1014S−1
Thus, frequency of yellow light emitted from sodium lamp
= 5.17×1014S−15.17×1014S−1
Wave number of yellow light, υ−=1λ=1580×10−9m=1.72×106m−1 υ−=1λ=1580×10−9m=1.72×106m−1

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6. (i). Find energy of each of the photons which correspond to light of frequency 3×1015Hz3×1015Hz

Ans: Energy (E) of a photon is given by the expression,
E=hυ
E=hυ
Where, h = Planck’s constant = 6.626×10−34Js6.626×10−34Js
υυ = frequency of light = 3×1015Hz3×1015Hz
Substituting the values in the given expression of Energy, E=(6.626×10−34)(3×1015)=1.988×10−18J

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7. Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0×10−10s2.0×10−10s

Ans: Frequency (υυ) of light = 1period=12×10−10s=5×109s−11period=12×10−10s=5×109s−1
Wavelength of light, λ=cυ
λ=cυ
c = velocity of light in vacuum = 3×108m/s3×108m/s
Substitution the value in the given expression λλ ,
λ=3×1085×109=6.0×10−2m
λ=3×1085×109=6.0×10−2m
Wave number of light, υ−=1λ=16.0×10−2=16.66m

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8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Ans: Energy (E) of a photon is given by the expression,
E=hυ
E=hυ
Energy of ‘n’ photons,
En=nhυ
En=nhυ
⇒n=Enλhc
⇒n=Enλhc
Where, λλ = wavelength of yellow light = 4000pm=4000×10−12m4000pm=4000×10−12m
c = velocity of light in vacuum = 3×108m/s3×108m/s
h = Planck’s constant = 6.626×10−34Js6.626×10−34Js
Substituting the values in the given expression ‘n’,
n=(1)×(4000×10−12)(6.626×10−34)(3×108)=2.012×1016
n=(1)×(4000×10−12)(6.626×10−34)(3×108)=2.012×1016
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012×10162.012×1016

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9. A photon of wavelength 4×107m4×107m strikes on metal surface, the work function of the metal being 2.13eV.
(i). Calculate the energy of photon (eV)

Ans: Energy of photon, E=hυ=hcλE=hυ=hcλ
Where, c = velocity of light in vacuum = 3×108m/s3×108m/s
h = Planck’s constant = 6.626×10−34Js6.626×10−34Js
λλ = wavelength of yellow light = 4×10−7m4×10−7m
Substituting the values in the given expression of E
, E=(6.626×10−34)(3×108)4×10−7=4.9695×10−19J
E=(6.626×10−34)(3×108)4×10−7=4.9695×10−19J
Hence, the energy of photon is 4.9695×10−19J

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(ii). Calculate the kinetic energy of the emission.

Ans: the kinetic energy of emission is given by,
Ek=hυ−hυ0=(E−W)eV
Ek=hυ−hυ0=(E−W)eV
=[4.9696×10−191.6020×10−19]−2.13eV=0.9720eV
=[4.9696×10−191.6020×10−19]−2.13eV=0.9720eV
Hence, the kinetic energy of emission is 0.97eV

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(iv). Calculate the velocity of the photoelectron (1eV=1.6020×10−19J1eV=1.6020×10−19J )

Ans: The velocity of a photoelectron (ν) can be calculated by the expression,
12mv2=hυ−hυ0
12mv2=hυ−hυ0
⇒v=2(hυ−hυ0)m−−−−−−−−−−√
⇒v=2(hυ−hυ0)m
Where, hυ−hυ0hυ−hυ0 is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:
v=2(0.9720×1.6020×10−19)9.10939×10−31kgJ−−−−−−−−−−−−−−−−−−−−−−−√
v=2(0.9720×1.6020×10−19)9.10939×10−31kgJ
⇒v=0.3418×1012m2s−2−−−−−−−−−−−−−−−−√=5.84×105ms−1
⇒v=0.3418×1012m2s−2=5.84×105ms−1
Hence, the velocity of the photoelectron is 5.84×105ms−15.84×105ms−1

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10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in kJmol−1kJmol−1

Ans: Energy of sodium, E=NAhcλE=NAhcλ
E=(6.023×1023mol−1)(6.626×10−34Js)(3×108ms−1)242×10−9m=4.947×105Jmol−1
E=(6.023×1023mol−1)(6.626×10−34Js)(3×108ms−1)242×10−9m=4.947×105Jmol−1
Hence, the ionization energy of sodium is 494kJmol−1494kJmol−1

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11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm0.57μm . Calculate the rate of emission of quanta per second.

Ans: power of Bulb, P = 25 Watt = 25Js−125Js−1
Energy of one photon, E=hυ=hcλE=hυ=hcλ
Substituting the values in the given expression of E,
E=(6.626×10−34)(3×108)0.57×10−6=34.87×10−20J
E=(6.626×10−34)(3×108)0.57×10−6=34.87×10−20J
Rate of emission of quanta per second,
=2534.87×10−20=7.169×1019s−1

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12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength6800A06800A0. Calculate threshold frequency (υ0υ0) and the work function (w0w0) of the metal.

Ans: Threshold wavelength of radian (λ0λ0) = 6800A0=6800×1010m6800A0=6800×1010m
Threshold frequency of metal (υ0υ0) = cλ0=3×108ms−16.8×10−7m=4.41×1014s−1cλ0=3×108ms−16.8×10−7m=4.41×1014s−1
Thus, threshold frequency of the metal is 4.41×1014s−14.41×1014s−1
Hence, the work function of the metal is,
w0=hv0=(6.626×10−34Js)(4.41×1014s−1)=2.922×10−19J

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13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n =2?

Ans: The ni=4ni=4 to nf=2nf=2 transition will rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,
E=2.18×10−18[1ni−1nf]
E=2.18×10−18[1ni−1nf]
Substituting the values in the given expression of E,
E=2.18×10−18[1ni2−1n2f]=2.18×10−18[142−122]=−4.0875×10−19J
E=2.18×10−18[1ni2−1n2f]=2.18×10−18[142−122]=−4.0875×10−19J
The negative sign indicates the energy of emission.
Wavelength of light emitted, λ=hcE
λ=hcE
Substituting the values in the given expression of λ λ
: λ=(6.626×10−34)(3×108)4.0875×10−19
λ=(6.626×10−34)(3×108)4.0875×10−19
λ=4.8631×10−7m
λ=4.8631×10−7m
λ=486nm
λ=486nm

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14. How much energy is required to ionize a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of the H atom (energy required to remove the electron from n =1 orbit).

Ans: The expression of energy is given by,
En=−(2.18×10−18)Z2n2
En=−(2.18×10−18)Z2n2
Where, Z = atomic number of atom
n = principal quantum number
For ionization from n1=5n1=5 to n2=∞n2=∞
ΔE=E∞−E5
ΔE=E∞−E5
⇒ΔE=[((−2.18×10−18J)(12)∞2)−((−2.18×10−18J)(12)52)]=2.18×10−18J
⇒ΔE=[((−2.18×10−18J)(12)∞2)−((−2.18×10−18J)(12)52)]=2.18×10−18J
Hence, the energy required for ionization from n1=5
n1=5
to n2=∞
n2=∞
is 8.72×1020J
8.72×1020J
The energy required for ionization from n1=1
n1=1
to n2=∞
n2=∞
ΔE=E∞−E1
ΔE=E∞−E1
⇒ΔE=[((−2.18×10−18J)(12)∞2)−((−2.18×10−18J)(12)12)]=2.18×10−18J
⇒ΔE=[((−2.18×10−18J)(12)∞2)−((−2.18×10−18J)(12)12)]=2.18×10−18J
Hence, less energy is required to ionize an electron in the 5th5th orbital of a hydrogen atom as compared to that in the ground state.

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15. What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Ans: When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:
(Image will be uploaded soon)
Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum. The number of spectral lines produced when an electron in the nthnth level drops down to the ground state is given by. n(n−1)2n(n−1)2
Number of spectral lines for n=6, 6(6−1)2=156(6−1)2=15

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16. (i). The energy associated with the first orbit in the hydrogen atom is −2.18×1018J−2.18×1018J&atom−1atom−1. What is the energy associated with the fifth orbit?

Ans: Energy associated with the fifth orbit of hydrogen atom is calculated as:
E5=−(2.18×10−18)(5)2=−2.18×10−1825
E5=−(2.18×10−18)(5)2=−2.18×10−1825
E5=−8.72×10−20J
(ii). Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Ans: Radius of Bohr’s nthnth orbit for hydrogen atom is given by,
rn=(0.0529nm)n2
rn=(0.0529nm)n2
For n = 5,
r5=(0.0529nm)×52
r5=(0.0529nm)×52
r5=1.3225nm
r5=1.3225nm

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17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans: For the Balmer series, ni=2ni=2 . Thus, the expression of wave number (υ−υ−) is given by
υ−=(1(2)2−1n2f)(1.097×107m−1)
υ−=(1(2)2−1n2f)(1.097×107m−1)
Wave number (υ−υ−) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, (υ−υ−) has to be the smallest.
For (υ−υ−) to be minimum, nfnf should be minimum. For the Balmer series, a transition from nini = 2 to nfnf= 3 is allowed. Hence, taking nfnf = 3, we get:
υ−=(1(2)2−1(3)2)(1.097×107m−1)
υ−=(1(2)2−1(3)2)(1.097×107m−1)
υ−=(14−19)(1.097×107m−1)
υ−=(14−19)(1.097×107m−1)
υ−=1.5236×106m−1
υ−=1.5236×106m−1

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18. What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is

−2.18×10−11ergs−2.18×10−11ergs
Ans: Energy (E) of the nthnth Bohr orbit of an atom is given by,
En=−(2.18×10−18)Z2n2
En=−(2.18×10−18)Z2n2
Where, Z = atomic number of the atom
Ground state energy= −2.18×10−11ergs=−2.18×10−11×10−7J=−2.18×10−18J−2.18×10−11ergs=−2.18×10−11×10−7J=−2.18×10−18J
Energy required to shift the electron from n = 1 to n = 5 is given as:
ΔE=E5−E1
ΔE=E5−E1
=−(2.18×10−18)(1)2(5)2−(−2.18×10−18)
=−(2.18×10−18)(1)2(5)2−(−2.18×10−18)
=(2.18×10−18)(2425)=2.0928×10−18
=(2.18×10−18)(2425)=2.0928×10−18
Weight of emitted light = hcEhcE
=(6.626×10−34)(3×108)(2.0928×10−18)=9.498×10−8m
=(6.626×10−34)(3×108)(2.0928×10−18)=9.498×10−8m

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19. The electron energy in hydrogen atom is given by En=(−2.18×10−18)n2JEn=(−2.18×10−18)n2J Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Ans: En=(−2.18×10−18)n2JEn=(−2.18×10−18)n2J
Given,
ΔE=E∞−E2
ΔE=E∞−E2
ΔE=((−2.18×10−18)∞2)−((−2.18×10−18)22)=0.545×10−18J
ΔE=((−2.18×10−18)∞2)−((−2.18×10−18)22)=0.545×10−18J
Energy required for ionization from n = 2 is given by, 0.545×10−18J 0.545×10−18J
ΔE=5.45×10−19J
ΔE=5.45×10−19J
λ=hcΔE
λ=hcΔE
, here λλ is the longest wavelength causing the transition.
λ=(6.626×10−34)(3×108)5.45×10−19=3.647×10−7m
λ=(6.626×10−34)(3×108)5.45×10−19=3.647×10−7m
λ=3647A0
λ=3647A0

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20. Calculate the wavelength of an electron moving with a velocity of 2.05×107ms−12.05×107ms−1

Ans: According to de Broglie’s equation, λ=hmvλ=hmv
Where,λλ = wavelength of moving particle
m = mass of particle, v = velocity of particle, h = Planck’s constant
Substituting the values in the expression of λλ:
λ=6.626×10−34Js(9.10939×10−31kg)(2.05×107ms−1)=3.548×10−11m
λ=6.626×10−34Js(9.10939×10−31kg)(2.05×107ms−1)=3.548×10−11m
Hence, the wavelength of an electron moving with a velocity of 2.05×107ms−12.05×107ms−1 is 3.548×10−11m3.548×10−11m

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21. The mass of an electron is 9.1×10−31kg9.1×10−31kg . If it is K.E. is 3.0×10−25J3.0×10−25J . Calculate its wavelengt

Ans: According to de Broglie’s equation, λ=hmvλ=hmv
Given, the K.E. of electrons is3.0×10−25J3.0×10−25J.
Since, K.E=12mv2K.E=12mv2
Velocity (v) = 2K.Em−−−−√2K.Em
⇒v=2(3.0×10−25J)9.10939×10−31kg=−−−−−−−−−−−−−−−−−√6.5866×104−−−−−−−−−−√ ⇒v=2(3.0×10−25J)9.10939×10−31kg=6.5866×104
V= 811.579ms−1811.579ms−1
Substituting the value in the expression of λλ :
λ=6.626×10−34Js(9.10939×10−31kg)(811.579ms−1)
λ=6.626×10−34Js(9.10939×10−31kg)(811.579ms−1)
λ=8.9625×10−7m
λ=8.9625×10−7m

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22. Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+,K+,Mg+2,Ca+2,S−2,ArNa+,K+,Mg+2,Ca+2,S−2,Ar

Ans: Isoelectronic species have the same number of electrons.
Number of electrons in sodium (Na) = 11
Number of electrons in (Na+Na+) = 10
A positive charge denotes the loss of an electron.
Similarly,
Number of electrons in K+K+ = 18
Number of electrons in Mg+2Mg+2 = 10
Number of electrons in Ca+2Ca+2 = 18
A negative charge denotes the gain of an electron by a species.
Number of electrons in sulphur (S) = 16
∴∴ Number of electrons in S−2S−2 = 18
Number of electrons in argon (Ar) = 18
Hence, the following are isoelectronic species:
(1) Na+Na+and Mg+2Mg+2 (10 electrons each)
(2) K+K+,Ca+2Ca+2, S−2S−2and Ar (18 electrons each).

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23. (i). Write the Electronic Configurations of the Following ions:

(a) H−H− ion
Ans: The electronic configuration of H atom is 1s11s1
A negative charge on the species indicates the gain of an electron by it.
∴∴ Electronic configuration of H−H− is 1s21s2
(b) Na+Na+ion
Ans: The electronic configuration of Na atom is 1s22s22p63s11s22s22p63s1
A positive charge on the species indicates the loss of an electron by it.
∴∴Electronic configuration of Na+Na+is 1s22s22p63s0or1s22s22p61s22s22p63s0or1s22s22p6
(c) O−2O−2 ion
Ans: The electronic configuration of O atom is 1s22s22p41s22s22p4
A negative charge on the species indicates the gain of two electrons by it.
∴∴Electronic configuration of O−2O−2is 1s22s22p61s22s22p6
(d) F−F− ion
Ans: The electronic configuration of F atom is 1s22s22p51s22s22p5
A negative charge on the species indicates the gain of an electron by it.
∴∴Electronic configuration of F−F−is 1s22s22p6

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(ii). What are the Atomic Numbers of Elements Whose Outermost Electrons are Represented by

(a) 3s13s1
Ans: Completing the electron configuration of the element as 1s22s22p63s11s22s22p63s1
∴∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴∴Atomic number of the element = 11
(b) 2p32p3
Ans: Completing the electron configuration of the element as 1s22s22p31s22s22p3
∴∴ Number of electrons present in the atom of the element
= 2+2+3=7
∴∴Atomic number of the element = 7
(c) 3p53p5
Ans: Completing the electron configuration of the element as 1s22s22p63s23p51s22s22p63s23p5
∴∴ Number of electrons present in the atom of the element
= 2+2+6+2+5=17
∴∴Atomic number of the element = 17
(iii).Which Atoms are Indicated by the Following Configurations?
(a) [He]2s1[He]2s1
Ans: The electronic configuration of the element is [He]2s1[He]2s1= 1s22s11s22s1
∴∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He]2s1[He]2s1is lithium (Li).
(b) [Ne]3s23p3[Ne]3s23p3
Ans: The electronic configuration of the element is [Ne]3s23p3[Ne]3s23p3= 1s22s22p63s23p31s22s22p63s23p3 .
∴∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne]3s23p3[Ne]3s23p3is phosphorus (P).
(c) [Ar]4s23d1[Ar]4s23d1
Ans: The electronic configuration of the element is [Ar]4s23d1[Ar]4s23d1=1s22s22p63s23p64s23d11s22s22p63s23p64s23d1 .
∴∴Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar]4s23d1[Ar]4s23d1is scandium (Sc).

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24. What is the Lowest Value of n That Allows G Orbitals To Exist?

Ans: For g-orbitals, l = 4. As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1). ∴∴ For l = 4, minimum value of n = 5.

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25. An Electron Is in One of the 3D Orbitals. Give the Possible Values of n, l and m, for This Electron.

Ans: For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (mlml) = –2, –1, 0, 1, 2

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26. (i). An Atom of an Element Contains 29 Electrons and 35 Neutrons. Deduce the Number of Protons.

Ans: For an atom to be neutral, the number of protons is equal to the number of electrons. ∴∴ Number of protons in the atom of the given element = 29
(II). An Atom of an Element Contains 29 Electrons and 35 Neutrons. Deduce the Electronic Configuration of the Given Element.
Ans: The electronic configuration of the atom is 1s22s23s23p64s23d101s22s23s23p64s23d10
The name of the element is Copper 3529Cu2935Cu

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27. Give the Number of Electrons in the Species H2+H2+ , H2H2 and O2+O2+

Ans: H2+H2+:
Number of electrons present in hydrogen molecule (H2H2) = 1 + 1 = 2
∴∴Number of electrons in H2+H2+= 2 – 1 = 1
H2H2:
Number of electrons inH2H2 = 1 + 1 = 2
O2+O2+:
Number of electrons present in oxygen molecule (O2O2) = 8 + 8 = 16
∴∴ Number of electrons in O2+O2+= 16-1 = 15

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28. (i). An atomic orbital has n = 3. What are the possible values of l and mlml ?

Ans: n = 3 (Given)
For a given value of n, l can have values from 0 to (n – 1).
∴∴For n = 3
l = 0, 1, 2
For a given value of l, ml can have (2l + 1) values.
∴∴For l = 0, m = 0
l = 1, m = –1, 0, 1
l = 2, m = –2, –1, 0, 1, 2
∴∴For n = 3
l = 0, 1, 2
m0m0 = 0
m1m1 = –1, 0, 1
m2m2 = –2, –1, 0, 1, 2
(ii). List the quantum numbers (mlml and l) of electrons for 3d orbital.
Ans: For 3d orbital, l = 2.
For a given value of l, mlml can have (2l + 1) values i.e., 5 values.
∴∴For l = 2
m2m2 = –2, –1, 0, 1, 2
(iii). Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Ans: Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1.
For a given value of n, l can have values from zero to (n – 1).
Therefore for l is equal to 1, the minimum value of n is 2.
Similarly,
For f-orbital, l = 4.
For l = 4, the minimum value of n is 5. Hence, 1p and 3f do not exist.

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29. Using s, p, d notations, describes the orbital with the following quantum numbers.

a) n = 1, l = 0; Ans: n = 1, l = 0 (Given) the orbital is 1s. b) n = 3; l =1 Ans: For n = 3 and l = 1 the orbital is 3p. c) n = 4; l = 2; Ans: For n = 4 and l = 2 the orbital is 4d. d) n = 4; l =3. Ans: For n = 4 and l = 3 the orbital is 4f.

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30. Explain, Giving Reasons, Which of the Following Sets of Quantum Numbers are Not

A	n = 0	l = 0	ml = 0	ms=+12
B	n = 1	l = 0	ml = 0	ms=−12
C	n = 1	l = 1	ml = 0	ms=+12
D	n = 2	l = 1	ml = 0	ms=−12
E	n = 3	l = 3	ml = - 3	ms=+12
F	n = 3	l = 1	ml = 0	ms=+12

Ans:
a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.
b) The given set of quantum numbers is possible.
c) The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.
d) The given set of quantum numbers is possible.
e) The given set of quantum numbers is not possible. For n = 3,
l = 0 to (3 – 1)
l = 0 to 2 i.e., 0, 1, 2
f) The given set of quantum numbers is possible.

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31. How Many Electrons in an Atom May Have the Following Quantum Numbers?
a) n = 4, and ms=−12ms=−12

Ans:
Total number of electrons in an atom for a value of n = 2n22n2
∴∴ For n = 4,
Total number of electrons = 2(4)2=322(4)2=32
The given element has a fully filled orbital as 1s22s23s23p64s23d101s22s23s23p64s23d10
Hence, all the electrons are paired.
∴∴ Number of electrons (having n = 4 and ms=−12ms=−12 ) = 16
b) n = 3, l = 0
Ans: n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

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32. Show That the Circumference of the Bohr Orbit for the Hydrogen Atom is an Integral Multiple of the De Broglie Wavelength Associated With the Electron Revolving Around the Orbit.

Ans: Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:
mvr=nh2πmvr=nh2π ----------- (i)
Where, n = 1, 2, 3 …
According to de Broglie’s equation:
λ=hmvλ=hmv Or mv=hλmv=hλ ------- (ii)
Substituting the value of ‘mv’ from expression (ii) in expression (i):
hrλ=nh2πhrλ=nh2π ----------------- (iii)
Since 2πr2πr represents the circumference of the Bohr orbit (r), it is proved by equation (iii) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

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33. What Transition in the Hydrogen Spectrum Would have the Same Wavelength as the Balmer Transition n = 4 to n = 2 of He+He+ spectrum

Ans: for He+He+ion, the wave number v−v− associated with the Balmer transition, n=4 to n =2 given by:
v−=1λ=RZ2(1n12−1n22)
v−=1λ=RZ2(1n12−1n22)
Where, n1=2andn2=4n1=2andn2=4 , Z = atomic number of Helium
v−=1λ=R22(122−142)=3R4
v−=1λ=R22(122−142)=3R4
⇒λ=43R
⇒λ=43R
According to the question, the desired transition for hydrogen will have the same wavelength as that of He+He+.
R(1)2(1n12−1n22)=3R4
R(1)2(1n12−1n22)=3R4
(1n12−1n22)=34
(1n12−1n22)=34
-------- (1)
By hit and trial method, the equality given by equation (1) is true only when n1=1n1=1 and n2=2n2=2
The transition for n2=2n2=2to n1=1n1=1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He+He+spectrum

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34. Calculate the energy required for the process He+(g)→He(g)2++e−He+(g)→He(g)2++e−
The ionization energy for the H atom in the ground state is 2.18×10−18Jatom−12.18×10−18Jatom−1 .

Ans: Energy associated with hydrogen-like species is given by,
En=−(2.18×10−18)Z2n2
En=−(2.18×10−18)Z2n2
For ground state of hydrogen atom,
ΔE=E∞−E1
ΔE=E∞−E1
⇒ΔE=0−[−2.18×10−18((1)2(1)2)]J
⇒ΔE=0−[−2.18×10−18((1)2(1)2)]J

∴ΔE=2.18×10−28J
∴ΔE=2.18×10−28J
For the given process, He+(g)→He(g)2++e−He+(g)→He(g)2++e−
An electron is removed from n = 1 to n = ∞.
ΔE=E∞−E1
ΔE=E∞−E1
⇒ΔE=0−[−2.18×10−18((2)2(1)2)]J
⇒ΔE=0−[−2.18×10−18((2)2(1)2)]J
∴ΔE=8.72×10−18J
∴ΔE=8.72×10−18J
The energy required for the process 8.72×10−18J8.72×10−18J

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35. If the Diameter of a Carbon Atom Is 0.15 nm, Calculate the Number of Carbon Atoms Which Can Be Placed Side by Side in a Straight Line Across a Length of Scale of Length 20 cm Long.

Ans: 1 m = 100 cm
1cm = 10−2m10−2m
Length of the scale = 20 cm = 20×10−2m20×10−2m
Diameter of a carbon atom = 0.15 nm = 0.15×10−9m0.15×10−9m
One carbon atom occupies 0.15×10−9m0.15×10−9m
Number of carbon atoms that can be placed in a straight line = 20×10−2m0.15×10−9m=1.33×109

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36. 2×1082×108 atoms of carbon are arranged side by side. Calculate the radius of the carbon atom if the length of this arrangement is 2.4 cm.

ns: Length of the given arrangement = 2.4 cm
Number of carbon atoms present = 2×1082×108
∴∴ Diameter of carbon atom = =2.4×10−22×108=1.2×10−16m=2.4×10−22×108=1.2×10−16m
Radius of carbon atom = diameter2=1.2×10−10m2=6×10−11mdiameter2=1.2×10−10m2=6×10−11m

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37. The diameter of the zinc atom is 2.6A02.6A0 .
a) Calculate radius of zinc atom in pm

Ans: Radius of zinc atom = diameter2=2.6A02=1.3×10−11m=13×10−12m=13pmdiameter2=2.6A02=1.3×10−11m=13×10−12m=13pm
b) Calculate the number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Ans: Length of the arrangement = 1.6 cm = 1.6×10−2m1.6×10−2m
Diameter of zinc atom = 2.6A02.6A0= 2.6×10−10m2.6×10−10m
Number of zinc atoms present in the arrangement =
1.6×10−2m2.6×10−10m=0.6153×108m
1.6×10−2m2.6×10−10m=0.6153×108m

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38. A certain particle carries 2.5×10−16C2.5×10−16C of static electric charge. Calculate the number of electrons present in it.

Ans: Charge on one electron = 1.6022×10−19C1.6022×10−19C
⇒1.6022×10−19C
⇒1.6022×10−19C
charge is carried by 1 electron.
Number of electrons carrying a charge of 2.5×10−16C2.5×10−16C
=1(2.5×10−16C1.6022×10−19)=1.560×103C=1560C
=1(2.5×10−16C1.6022×10−19)=1.560×103C=1560C

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39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is −1.282×10−18C−1.282×10−18C , calculate the number of electrons present in it.

Ans: Charge on the oil drop = −1.282×10−18C−1.282×10−18C
Charge on one electron = 1.6022×10−19C1.6022×10−19C
∴∴ Number of electrons present on the oil drop = =1.282×10−18C1.6022×10−19C=0.8001×101=8.0=1.282×10−18C1.6022×10−19C=0.8001×101=8.0

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40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the αα-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Ans: A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.
Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of αα-particles (positively charged).

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41. Symbols 7935Br3579Br and 79Br79Br can be written. Whereas symbols 3579Br7935Br and 35Br35Brare not acceptable. Answer briefly.

Ans: The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is AZXZAX
Hence, 7935Br3579Bris acceptable, but 3579Br7935Bris not acceptable.
79Br79Br can be written but 35Br35Br cannot be written because the atomic number of an element is constant, but the atomic mass of an element
depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.

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42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans: Let the number of protons in the element be x.
∴∴Number of neutrons in the element = x+0.317x=1.317xx+0.317x=1.317x
According to the question,
Mass number of the element = 81
(Number of protons + number of neutrons) = 81
⇒x+1.317x=81
⇒x+1.317x=81
2.317x=81
2.317x=81
x=812.317=35
x=812.317=35
Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
The atomic symbol of the element is 8135Br3581Br

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43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Ans: Let the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present = =x+0.111x=1.111x=x+0.111x=1.111x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
∴∴ Number of protons in the neutral atom = x – 1
Given,
Mass number of the ion = 37
∴(x−1)+1.111x=37
∴(x−1)+1.111x=37
2.111x=38
2.111x=38
X = 18
Number of electrons = 18; Number of protons = 18 – 1 = 17
Atomic number of the ion = 17; Atom correspondence to ion = ClCl ∴∴The symbol of the ion is 1737Cl−1

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44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion

Ans: Let the number of electrons present in ion A+3A+3 be ‘x’
Number of neutrons in it = x+0.304x=1.304xx+0.304x=1.304x
Since the ion is tripositive,
Number of electrons in neutral atom = x + 3
∴∴Number of protons in neutral atom = x + 3
Given,
Mass number of the ion = 56
(x + 3) + (1.304x) = 56
2.304x = 53
x= 23
Number of protons = x + 3 = 23 + 3 = 26 The symbol of the ion 2656Fe3+

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45. Arrange the Following Type of Radiations in Increasing Order of Frequency:
radiation from microwave oven
amber light from traffic signal
radiation from FM radio
cosmic rays from outer space and
X-rays.

Ans: The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays
The increasing order of wavelength is as follows:
Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio.

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46. Nitrogen laser produces radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6×10245.6×1024 , calculate the power of the laser.

Ans: Power of laser = Energy with which it emits photons
Power = E=NhcλE=Nhcλ
Where, N = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λλ= wavelength of radiation
Substituting the values in the given expression of E,
E=(5.6×1024)(6.626×10−34Js)(3×108ms−1)(337.1×10−9m)=3.33×106J
E=(5.6×1024)(6.626×10−34Js)(3×108ms−1)(337.1×10−9m)=3.33×106J
Hence, the power of laser is 3.33×106J3.33×106J

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47. Neon Gas is Generally Used in Sign Boards. If it Emits Strongly at 616 nm,
a) Calculate the frequency of emission.

Ans: Wavelength of radiation emitted = 616 nm = 616×10−9m616×10−9m (Given)
Frequency of emission, v=cλv=cλ
Where, c = velocity of radiation
λ=λ= Wavelength of radiation
Substituting the values in the given expression of frequency of emission,
v=3.0×108m/s616×10−9m=4.87×1014s−1
v=3.0×108m/s616×10−9m=4.87×1014s−1
Hence, Frequency of emission, v=4.87×1014s−1v=4.87×1014s−1
b) Calculate Distance Traveled by this Radiation in 30 s
Ans: velocity of radiation, c = 3×108m/s3×108m/s
Distance travelled by this radiation in 30 s = (3×108ms−1)(90s)=9×109m(3×108ms−1)(90s)=9×109m
c) Calculate energy of quantum…
Ans: energy of Quantum, E=hv=(6.626×10−34Js)(4.87×1014s−1)=32.27×10−20JE=hv=(6.626×10−34Js)(4.87×1014s−1)=32.27×10−20J
Energy of Quantum (E) = 32.27×10−20J32.27×10−20J
d) Calculate number of quanta present if it produces 2 J of energy.
Ans: Energy of one photon (E) = 32.27×10−20J32.27×10−20J
Therefore, 32.27×10−20J32.27×10−20J of energy is present in 1 quantum number of quanta in 2J of energy = 2J32.27×10−20J=6.19×10182J32.27×10−20J=6.19×1018

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48. In astronomical observations, signals observed from distant stars are generally weak. If the photon detector receives a total of 3.15×10−183.15×10−18J from the radiations of 600 nm, calculate the number of photons received by the detector.

Ans: From the expression of energy of one photon, E=hcλE=hcλ
Where, λλ = wavelength of radiation
h = Planck’s constant
c= velocity of radiation
Substituting the values in the given expression of E:
E=(6.626×10−34Js)(3×108ms−1)(600×10−9m)
E=(6.626×10−34Js)(3×108ms−1)(600×10−9m)
Energy of one photon, E = 3.313×10−19J3.313×10−19J
Number of photons with received with 3.15×10−183.15×10−18J energy,
=3.15×10−18J3.313×10−19J=9.5≈10
=3.15×10−18J3.313×10−19J=9.5≈10
. Hence, the number of photons received by the detector = 10

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49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nanosecond range. If the radiation source has a duration of 2 ns and the number of photons emitted during the pulse source is 2.5×10152.5×1015 , calculate the energy of the source.

Ans: frequency of radiation, v=12.0×10−9s=5×108s−1v=12.0×10−9s=5×108s−1
Energy of source, E=NhvE=Nhv
Where, N = number of photons emitted
h = Planck’s constant
v = frequency of radiation
Substituting the values in the given expression of E,>
\[E=(2.5\times {{10}^{15}})(6.626\times {{10}^{-34}}Js)(5.0\times
{{10}^{8}}{{s}^{-1}})=8.282\times {{10}^{-10}}J\] Hence, the energy of source, E=8.282×10−10J

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50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Ans: Given
Wavelength associated with first transition, λ1=589nm=589×10−9mλ1=589nm=589×10−9m
Wavelength associated with second transition, λ2=589.6nm=589.6×10−9mλ2=589.6nm=589.6×10−9m
Frequency of first wavelength is, v1=cλ1=3×108ms−1589×10−9m=5.093×1014s−1v1=cλ1=3×108ms−1589×10−9m=5.093×1014s−1
And, frequency of second wavelength is, v2=cλ2=3×108ms−1589.6×10−9m=5.088×1014s−1v2=cλ2=3×108ms−1589.6×10−9m=5.088×1014s−1
Energy difference between two excited states is given as,
ΔE=hv1−hv2=h(v1−v2)
ΔE=hv1−hv2=h(v1−v2)
⇒ΔE=(6.626×10−34Js)(5.093×1014s−1−5.088×1014s−1)=3.31×10−22J

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51. The Work Function for Cesium Atoms is 1.9 eV.
a) Calculate the threshold wavelength

Ans: It is given that the work function (W0W0) for caesium atom is 1.9 eV.
From the expression, Wo=hcλ0Wo=hcλ0
We get, λo=hcW0λo=hcW0
Where, λ0λ0 = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression ofλ0λ0:
λ0=(6.626×10−1Js)(3.0×108ms−1)1.9×1.602×10−19J=6.53×10−7mλ0=(6.626×10−1Js)(3.0×108ms−1)1.9×1.602×10−19J=6.53×10−7m
Hence, threshold wavelength λ0λ0is 653 nm
b) Calculate the threshold frequency of the radiation.
Ans: From the expression, W0=hv0W0=hv0
we get, v0=hW0v0=hW0
Where, v0v0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression ofv0v0,
v0=1.9×1.602×10−19J6.626×10−34Js=4.593×1014s−1v0=1.9×1.602×10−19J6.626×10−34Js=4.593×1014s−1
Hence, the threshold frequency of the radiation, v0=4.593×1014s−1v0=4.593×1014s−1
c) If the cesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Ans: According to the question:
Wavelength used in irradiation, λλ = 500nm
Kinetic energy = h(v−v0)=hc(1λ−1λ0)h(v−v0)=hc(1λ−1λ0)
⇒K.E=(6.626×10−34Js)(3.0×108ms−1)(1500×10−9m−1653×10−9m)
⇒K.E=(6.626×10−34Js)(3.0×108ms−1)(1500×10−9m−1653×10−9m)
=(1.9878×10−26)(153×109)653×500=9.3149×10−20J
=(1.9878×10−26)(153×109)653×500=9.3149×10−20J
Kinetic energy of the ejected photoelectron = 9.3149×10−20J9.3149×10−20J
Since, kinetic energy = 12mv2=9.3149×10−20J12mv2=9.3149×10−20J
v=2(9.3149×10−20J)9.10939×1011m2s−2−−−−−−−−−−−−−−−−−√=4.52×105ms−1
v=2(9.3149×10−20J)9.10939×1011m2s−2=4.52×105ms−1
Hence, the velocity of the ejected photoelectron (v) is 4.52×105ms−1

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52. Following Results are Observed When Sodium Metal is Irradiated With Different Wavelengths.

λ(nm)	500	450	400
v×105(cms−1)	2.55	4.35	5.35

Ans: Assuming the threshold wavelength to be λ0nmλ0nm , the kinetic energy of the radiation is given as:
h(v−v0)=12mv2
h(v−v0)=12mv2
Three different equalities can be formed by the given value as:
hc(1λ−1λ0)=12mv2
hc(1λ−1λ0)=12mv2
hc(1500×109−1λ0×10−9m)=12m(2.55×105×10−2ms−1)2
hc(1500×109−1λ0×10−9m)=12m(2.55×105×10−2ms−1)2
hc10−9m(1500−1λ0)=12m(2.55×103ms−1)2hc10−9m(1500−1λ0)=12m(2.55×103ms−1)2 ---------- (1)
Similarly,
hc10−9m(1450−1λ0)=12m(3.45×103ms−1)2hc10−9m(1450−1λ0)=12m(3.45×103ms−1)2 -------------- (2)
hc10−9m(1400−1λ0)=12m(5.35×103ms−1)2hc10−9m(1400−1λ0)=12m(5.35×103ms−1)2 --------------- (3)
Dividing equation (3) by equation (1):
[λ0−400400λ0][λ0−500500λ0]=(5.35×103ms−1)2(2.55×103ms−1)2
[λ0−400400λ0][λ0−500500λ0]=(5.35×103ms−1)2(2.55×103ms−1)2
5λ0−20004λ0−2000=(5.352.55)2=28.62256.5025
5λ0−20004λ0−2000=(5.352.55)2=28.62256.5025
5λ0−20004λ0−2000=4.40177
5λ0−20004λ0−2000=4.40177
17.6070λ0−5λ0=8803.537−2000
17.6070λ0−5λ0=8803.537−2000
λ0=6805.53712.607=539.8nm=540nm
λ0=6805.53712.607=539.8nm=540nm
Threshold wavelength = 540 nm
b) Calculate Planck’s constant
Ans: the question is not done due to the incorrect values of velocity given in the question.

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53. The Ejection of the Photoelectron From the Silver Metal in the Photoelectric Effect Experiment Can Be Stopped by Applying the Voltage of 0.35 v When the Radiation 256.7 nm Is Used. Calculate the Work Function for Silver Metal.

Ans: From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0W0) of radiation and its kinetic energy (K.E.) i.e., E=W0+K.EE=W0+K.E
we get, W0=E−K.EW0=E−K.E
Energy of incident photon, E=hcλE=hcλ
0 Where, λλ = wavelength of radiation
C = velocity of radiation
h = Planck’s constant
Substituting the values in the given expression of E:
E=(6.626×10−34Js)(3.0×108ms−1)256.7×10−9m=7.744×10−19J
E=(6.626×10−34Js)(3.0×108ms−1)256.7×10−9m=7.744×10−19J
E=7.744×10−191.602×10−19C=4.83eV
E=7.744×10−191.602×10−19C=4.83eV
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence, K.E = 0.35 eV
Work function, W0=E−K.EW0=E−K.E
W0=4.83eV−0.35eV=4.48eV
W0=4.83eV−0.35eV=4.48eV

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54. If the Photon of the Wavelength 150 PM Strikes an Atom and One of Its Inner Bound Electrons Is Ejected Out With a Velocity of 1.5×107ms−11.5×107ms−1 , Calculate the energy with which it is bound to the nucleus

Ans: Energy of incident photon (E) is given by, E=hcλE=hcλ
E=(6.626×10−34Js)(3.0×108ms−1)(150×10−12m)=13.252×10−16J
E=(6.626×10−34Js)(3.0×108ms−1)(150×10−12m)=13.252×10−16J
Energy of the electron ejected (K.E) = 12mev212mev2
=12(9.10939×10−31kg)(1.5×107ms−1)=1.025×10−16J
=12(9.10939×10−31kg)(1.5×107ms−1)=1.025×10−16J
Hence, the energy with which the electron is bound to the nucleus can be obtained as: = E – K.E
=13.252×10−16J−1.02×10−16J=12.227×10−16J13.252×10−16J−1.02×10−16J=12.227×10−16J
=12.227×10−161.602×10−19C=7.6×103eV
=12.227×10−161.602×10−19C=7.6×103eV
The energy with which it is bound to the nucleus is 7.6×103eV7.6×103eV

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55. Emission Transitions in the Paschen Series end at orbit n = 3 and start from orbit n and can be represented as v=3.29×105(Hz)[132−1n2]v=3.29×105(Hz)[132−1n2] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Ans: Wavelength of transition = 1285 nm = 1285×10−9m1285×10−9mv=3.29×105(Hz)[132−1n2]v=3.29×105(Hz)[132−1n2]
Since, v=cλ=3.0×108ms−11285×10−9m=2.33×1014s−1v=cλ=3.0×108ms−11285×10−9m=2.33×1014s−1
Substituting the value of v in the given expression,
2.33×1014=3.29×105(Hz)[132−1n2]2.33×1014=3.29×105(Hz)[132−1n2]
⇒(19−1n2)=2.33×10143.29×1015
⇒(19−1n2)=2.33×10143.29×1015
⇒1n2=1.1×10−1−0.7082×10−1
⇒1n2=1.1×10−1−0.7082×10−1
⇒1n2=4.029×10−2
⇒1n2=4.029×10−2
n=14.029×10−2−−−−−−−−−−−√=4.98≈5
n=14.029×10−2=4.98≈5
Hence, for the transition to be observed at 1285 nm, n = 5. The spectrum lies in the infra-red region.

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56. Calculate the Wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Ans: The radius of the nthnth orbit of hydrogen-like particles is given by,
r=0.529n2ZA0
r=0.529n2ZA0
r=52.9n2Zpm
r=52.9n2Zpm
For radius, r1=1.3225nm=1.3225×10−9m=1322.5pmr1=1.3225nm=1.3225×10−9m=1322.5pm
n12=r1Z52.9=1322.5Z52.9
n12=r1Z52.9=1322.5Z52.9
Similarly,
n22=211.6Z52.9
n22=211.6Z52.9
n12n22=1322.5211.6=6.25
n12n22=1322.5211.6=6.25
n1n2=2.5=2510=52
n1n2=2.5=2510=52
⇒n1=5&n2=2
⇒n1=5&n2=2
Thus, the transition is from the 5th5th orbit to the 2nd2nd orbit. It belongs to the Balmer series. Wave number (v−v−) for the transition is given by,
=1.097×107(122−152)
=1.097×107(122−152)
=1.097×107(21100)=2.303×106m−1
=1.097×107(21100)=2.303×106m−1
Wavelength associated with the emission transition is given by,
\ λ=1v−=12.303×106m−1=0.434×106m=434nm
λ=1v−=12.303×106m−1=0.434×106m=434nm

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57. Dual Behavior of Matter Proposed by De Broglie Led to the Discovery of Electron Microscopes Often Used for the Highly Magnified Images of Biological Molecules and Other Types of Material. If the velocity of the electron in this microscope is 1.6×106ms−11.6×106ms−1 , calculate the de Broglie wavelength associated with this electron.

Ans: From de Broglie’s equation,
λ=hmv=6.626×10−34Js(9.10939×10−31kg)(1.6×106ms−1)=4.55×10−10m=455pm
λ=hmv=6.626×10−34Js(9.10939×10−31kg)(1.6×106ms−1)=4.55×10−10m=455pm
de Broglie’s wavelength associated with the electron is 455 pm.

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58. Similar to Electron Diffraction, Neutron Diffraction Microscope is Also Used for the Determination of the Structure of Molecules. If the Wavelength Used Here is 800 PM, Calculate the Characteristic Velocity Associated With the Neutron

Ans: From de Broglie’s equation,
λ=hmv
λ=hmv
Then, v=hmλv=hmλ
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
λλ = wavelength
Substituting the values in the expression of velocity (v),
v=(6.626×10−34)Kgm2s−1(1.675×10−27)(8×10−10m)=6.626×1031.675×8=4.94×102ms−1=494ms−1
v=(6.626×10−34)Kgm2s−1(1.675×10−27)(8×10−10m)=6.626×1031.675×8=4.94×102ms−1=494ms−1
Velocity associated with the neutron = 494ms−1494ms−1

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59. If the velocity of the electron in Bohr’s first orbit is 2.19×106ms−12.19×106ms−1 calculate the de Broglie wavelength associated with it.

Ans: From de Broglie’s equation,
λ=hmv
λ=hmv
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
λλ = wavelength
Substituting the values in the expression of λ:
λ=(6.626×10−34Js)(9.10939×10−31kg)(2.19×106ms−1)=3.32×10−10m=332×10−12m=332pm
λ=(6.626×10−34Js)(9.10939×10−31kg)(2.19×106ms−1)=3.32×10−10m=332×10−12m=332pm
Wavelength associated with the electron = 332 pm

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60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×105ms−14.37×105ms−1 . If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Ans: From de Broglie’s equation,
λ=hmv
λ=hmv
Substituting the values in the expression,
λ=(6.626×10−34Js)(0.1kg)(4.37×105ms−1)=1.516×10−38m

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61. If the position of the electron is measured within an accuracy of +0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h4πm×0.05nmh4πm×0.05nm . Is there any problem in defining this value?

Ans: From Heisenberg’s uncertainty principle,
Δx×Δp=h4π⇒Δp=1Δxh4π
Δx×Δp=h4π⇒Δp=1Δxh4π
Where, ΔxΔx = uncertainty in position of the electron
ΔpΔp = uncertainty in momentum of the electron
Substituting the values in the expression ofΔpΔp,
Δp=10.002nm6.626×10−34Js4×3.14Δp=10.002nm6.626×10−34Js4×3.14
⇒Δp=2.637×10−23kgms−1⇒Δp=2.637×10−23kgms−1
Actual momentum = h4πm×0.05nmh4πm×0.05nm=
6.626×10−34Js4×3.14×5.0×10−11m=1.055×10−24kgms−16.626×10−34Js4×3.14×5.0×10−11m=1.055×10−24kgms−1
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

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62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
n = 4, l = 2, ml=−2,ms=−12ml=−2,ms=−12
n = 3, l = 2, ml=1,ms=+12ml=1,ms=+12
n = 4, l = 1, ml=0,ms=+12ml=0,ms=+12
n = 3, l = 2, ml=−2,ms=−12ml=−2,ms=−12
n = 3, l = 1, ml=−1,ms=+12ml=−1,ms=+12
n = 4, l = 1, ml=0,ms=+12ml=0,ms=+12

Ans: For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

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63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electrons experiences the lowest effective nuclear charge?

Ans: Nuclear charge experienced by an electron (present in a multi-electron atom) is dependent upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also
decreases.
Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atoms with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

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64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

(i) 2s and 3s
Ans: Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron(s) in it.
The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.
(ii) 4d and 4f
Ans: 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.
(iii) 3d and 3p
Ans: 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

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65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Ans: Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
The higher the atomic number, the higher is the nuclear charge.
Silicon has 14 protons while aluminium has 13 protons.
Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13).
Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

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66. Indicate the number of unpaired electrons a) Phosphorous (P)

Ans: Atomic number = 15
The electronic configuration of P is: 1s22s22p63s23p31s22s22p63s23p3
The orbital picture of P can be represented as:
(Image will be uploaded soon)
From the orbital picture, phosphorus has three unpaired electrons.
b) Silicon (Si)
Ans: Atomic number = 14
The electronic configuration of Si is: 1s22s22p63s23p21s22s22p63s23p2
The orbital picture of Si can be represented as:
(Image will be uploaded soon)
From the orbital picture, phosphorus has two unpaired electrons.
c) Chromium (Cr)
Ans: Atomic number = 24
The electronic configuration of Cr is: 1s22s22p63s23p64s13d51s22s22p63s23p64s13d5
The orbital picture of chromium is:
(Image will be uploaded soon)
From the orbital picture, chromium has six unpaired electrons.
d) Iron (Fe)
Ans: Atomic number = 26
The electronic configuration is: 1s22s22p63s23p64s23d61s22s22p63s23p64s23d6
The orbital picture of iron is:
(Image will be uploaded soon)
From the orbital picture, iron has four unpaired electrons.
e) Krypton (Kr)
Ans: Atomic number = 36
The electronic configuration is:1s22s22p63s23p64s23d104p61s22s22p63s23p64s23d104p6
The orbital picture of krypton is:
(Image will be uploaded soon) Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

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67. Answer the following. a) How many subshells are associated with n = 4?

Ans: n = 4 (Given)
For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).
Hence, l = 0, 1, 2, 3
Thus, four subshells are associated with n = 4, which are s, p, d and f.

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Last Updated on: Mar 28, 2024