{"id":121295,"date":"2022-05-17T14:45:27","date_gmt":"2022-05-17T09:15:27","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=121295"},"modified":"2022-05-17T14:45:27","modified_gmt":"2022-05-17T09:15:27","slug":"chapter-12-atoms-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-12-atoms-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 12 \u2013 Atoms Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"

1. Fill in the blanks using the given options:
\n1. The size of the atoms in Thomson’s model are _______ the atomic size in Rutherford’s model (much greater than\/no different from\/much lesser than).<\/h2>\n

Ans: The sizes of the atoms in Thomson’s model are no different from the atomic size in Rutherford’s model.<\/h3>\n

2. In the ground state of _______ electrons are in stable equilibrium, while in _______ electrons always experience a net force. (Thomson’s model\/Rutherford’s model)<\/h2>\n

Ans: In the ground state of Thomson’s model, the electrons are in stable equilibrium, while in Rutherford’s model, electrons always experience a net force.<\/h3>\n

3. A classical atom based on _______ is doomed to collapse. (Thomson’s model\/ Rutherford’s model.)<\/h2>\n

Ans: A classical atom based on Rutherford’s model is doomed to collapse.<\/h3>\n

4. An atom features a nearly continuous mass distribution in _______ but features a highly non- uniform mass distribution in _______ (Thomson’s model\/ Rutherford’s model.)<\/h2>\n

Ans: An atom features a nearly continuous mass distribution in Thomson’s model, but features a highly non-uniform mass distribution in Rutherford’s model.<\/h3>\n

5. The positively charged part of the atom possesses most of the mass in _______ (Rutherford’s model\/ Thomson’s model \/both the models.)<\/h2>\n

Ans: The positively charged part of the atom possesses most of the mass in both the models.<\/h3>\n

2. If you\u2019re given a chance to repeat the \u03b1\u2212\u03b1\u2212particle scattering experiment employing a thin sheet of solid hydrogen instead of the gold foil (Hydrogen is a solid at temperatures below 14 K). What results would you expect?<\/h2>\n

Ans: In the \u03b1\u2212\u03b1\u2212particle scattering experiment, when a thin sheet of solid hydrogen is replaced with the gold foil, the scattering angle would not turn out to be large enough.
\nThis is because the mass of hydrogen is smaller than the mass of incident \u03b1\u2212\u03b1\u2212particles. Also, the mass of the scattering particle is more than the target nucleus (hydrogen).
\nAs a consequence, the \u03b1\u2212\u03b1\u2212particles would not bounce back when solid hydrogen is utilized in the \u03b1\u2212\u03b1\u2212particle scattering experiment and hence, we cannot evaluate the size of the hydrogen nucleus.<\/h3>\n

3. What\u2019s the shortest wavelength present within the Paschen series of spectral lines?<\/h2>\n

Ans: We know the Rydberg\u2019s formula is given as;
\nhc\u03bb=21.76\u00d710\u221219[1n21\u22121n22]hc\u03bb=21.76\u00d710\u221219[1n12\u22121n22]
\nHere,
\nhh = Planck\u2019s constant = 6.6\u00d710\u221234Js6.6\u00d710\u221234Js
\ncc = speed of light = 3\u00d7108m\/s3\u00d7108m\/s
\n(n1n1 and n2n2 are integers)
\nThe shortest wavelength present within the Paschen series of the spectral lines is for values n1=3n1=3 and n2=\u221en2=\u221e.
\n\u21d2\u21d2 hc\u03bb=21.76\u00d710\u221219[132\u22121\u221e2]hc\u03bb=21.76\u00d710\u221219[132\u22121\u221e2]
\n\u21d2\u21d2 \u03bb=6.6\u00d710\u221234\u00d73\u00d7108\u00d7921.76\u00d710\u221219\u03bb=6.6\u00d710\u221234\u00d73\u00d7108\u00d7921.76\u00d710\u221219
\n\u21d2\u21d2 \u03bb=8.189\u00d710\u22127m\u03bb=8.189\u00d710\u22127m
\n\u21d2\u21d2 \u03bb=818.9nm\u03bb=818.9nm<\/h3>\n

4. The two energy levels in an atom are separated by a difference of 2.3eV2.3eV. What is the frequency of radiation emitted when the atom makes a transition from the higher level to the lower level?<\/h2>\n

Ans: Given that the distance between the two energy levels in an atom is E=2.3eVE=2.3eV.
\n\u21d2E=2.3\u00d71.6\u00d710\u221219\u21d2E=2.3\u00d71.6\u00d710\u221219
\n\u21d2E=3.68\u00d710\u221219J\u21d2E=3.68\u00d710\u221219J
\nLet \u03bd\u03bd be the frequency of radiation emitted when the atom jumps from the upper level to the lower level.
\nThe relation for energy is given as;
\nE=h\u03bdE=h\u03bd
\nHere,
\nh=h= Planck\u2019s constant =6.6\u00d710\u221234Js=6.6\u00d710\u221234Js
\n\u21d2\u03bd=Eh\u21d2\u03bd=Eh
\n\u21d2\u03bd=3.38\u00d710\u2212196.62\u00d710\u221232\u21d2\u03bd=3.38\u00d710\u2212196.62\u00d710\u221232
\n\u21d2\u03bd=5.55\u00d71014Hz\u21d2\u03bd=5.55\u00d71014Hz
\nClearly, the frequency of the radiation is 5.6\u00d71014Hz5.6\u00d71014Hz.<\/h3>\n

5. For a hydrogen atom, the ground state energy is \u221213.6eV\u221213.6eV . What are the kinetic and potential energies of the electron during this state?<\/h2>\n

Ans: Provided that the ground state energy of hydrogen atom, E=\u221213.6eVE=\u221213.6eVwhich is the total energy of a hydrogen atom.
\nHere, kinetic energy is equal to the negative of the total energy.
\nKinetic energy =\u2212E=\u2212(\u221213.6)=13.6eV=\u2212E=\u2212(\u221213.6)=13.6eV
\nThe potential energy is the same as the negative of two times kinetic energy.
\nPotential energy =\u22122\u00d7(13.6)=\u221227.2eV=\u22122\u00d7(13.6)=\u221227.2eV
\n\u2234\u2234 The kinetic energy of the electron is 13.6eV13.6eV and the potential energy is \u221227.2eV\u221227.2eV .<\/h3>\n

6. A hydrogen atom absorbs a photon when it is in the ground level, this excites it to the n=4n=4 level. Find out the wavelength and frequency of the photon.<\/h2>\n

Ans: It is known that for ground level absorption, n1=1n1=1
\nLet E1E1 be the energy of this level. It is known that E1E1 is related with n1n1 as;
\nE1=\u221213.6n21eVE1=\u221213.6n12eV
\n\u21d2E1=\u221213.612=\u221213.6eV\u21d2E1=\u221213.612=\u221213.6eV
\nWhen the atom jumps to a higher level, n2=4n2=4.
\nLet E2E2 be the energy of this level.
\n\u21d2E2=\u221213.6n22eV\u21d2E2=\u221213.6n22eV
\n\u21d2E2=\u221213.642=\u221213.616eV\u21d2E2=\u221213.642=\u221213.616eV
\nThe amount of energy absorbed by the photon is given as;
\nE=E1\u2212E2E=E1\u2212E2
\n\u21d2E=(\u221213.616)\u2212(\u221213.61)\u21d2E=(\u221213.616)\u2212(\u221213.61)
\n\u21d2E=13.6\u00d71516eV\u21d2E=13.6\u00d71516eV
\n\u21d2E=13.6\u00d71616\u00d71.6\u00d710\u221219\u21d2E=13.6\u00d71616\u00d71.6\u00d710\u221219
\n\u21d2E=2.04\u00d710\u221218J\u21d2E=2.04\u00d710\u221218J
\nFor a photon of wavelength \u03bb\u03bb , the expression of energy is written as;
\nE=hc\u03bbE=hc\u03bb
\nHere,
\nh=h= Planck\u2019s constant =6.6\u00d710\u221234Js=6.6\u00d710\u221234Js
\nc=c= speed of light =3\u00d7108m\/s=3\u00d7108m\/s
\n\u21d2\u03bb=hcE\u21d2\u03bb=hcE
\n\u21d2\u03bb=6.6\u00d710\u221234\u00d73\u00d71082.04\u00d710\u221218\u21d2\u03bb=6.6\u00d710\u221234\u00d73\u00d71082.04\u00d710\u221218
\n\u21d2\u03bb=9.7\u00d710\u22128m\u21d2\u03bb=9.7\u00d710\u22128m
\n\u21d2\u03bb=97nm\u21d2\u03bb=97nm
\nAlso, frequency of a photon is given by the relation,
\n\u03bd=c\u03bb\u03bd=c\u03bb
\n\u21d2\u03bd=3\u00d71089.7\u00d710\u22128\u22483.1\u00d71015Hz\u21d2\u03bd=3\u00d71089.7\u00d710\u22128\u22483.1\u00d71015Hz
\nClearly, the wavelength of the photon is 97nm whereas the frequency is 3.1\u00d71015Hz3.1\u00d71015Hz.<\/h3>\n

7. Answer the following questions.
\n1. Using the Bohr\u2019s model, calculate the speed of the electron in a hydrogen atom in the n=1,2n=1,2 and 33 levels.<\/h2>\n

Ans: Consider \u03bd1\u03bd1 to be the orbital speed of the electron in a hydrogen atom in the ground state level n1=1n1=1 . For charge (e)(e) of an electron, \u03bd1\u03bd1 is given by the relation,
\n\u03bd1=e2n14\u03c0\u22080(h2\u03c0)\u03bd1=e2n14\u03c0\u22080(h2\u03c0)
\n\u21d2\u03bd1=e22\u22080h\u21d2\u03bd1=e22\u22080h
\nHere,
\ne=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C
\n\u22080=\u22080=Permittivity of free space =8.85\u00d710\u221212N\u22121C2m\u22122=8.85\u00d710\u221212N\u22121C2m\u22122
\nh=h= Planck\u2019s constant =6.6\u00d710\u221234Js=6.6\u00d710\u221234Js
\n\u21d2\u03bd1=(1.6\u00d710\u221219)22\u00d78.85\u00d710\u221212\u00d76.62\u00d710\u221234\u21d2\u03bd1=(1.6\u00d710\u221219)22\u00d78.85\u00d710\u221212\u00d76.62\u00d710\u221234
\n\u21d2\u03bd1=0.0218\u00d7108\u21d2\u03bd1=0.0218\u00d7108
\n\u21d2\u03bd1=2.18\u00d7106m\/s\u21d2\u03bd1=2.18\u00d7106m\/s
\nFor level n2=2n2=2 , we can write the relation for the corresponding orbital speed as;
\n\u03bd2=e2n22\u22080h\u03bd2=e2n22\u22080h
\n\u21d2\u03bd2=(1.16\u00d710\u221219)22\u00d72\u00d78.85\u00d710\u221212\u00d76.62\u00d710\u221234\u21d2\u03bd2=(1.16\u00d710\u221219)22\u00d72\u00d78.85\u00d710\u221212\u00d76.62\u00d710\u221234
\n\u21d2\u03bd2=1.09\u00d7106m\/s\u21d2\u03bd2=1.09\u00d7106m\/s
\nAnd, for n3=3n3=3 , we can write the relation for the corresponding orbital speed as;
\n\u03bd3=e2n32\u22080h\u03bd3=e2n32\u22080h
\n\u21d2\u03bd3=(1.16\u00d710\u221219)23\u00d72\u00d78.85\u00d710\u221212\u00d76.62\u00d710\u221234\u21d2\u03bd3=(1.16\u00d710\u221219)23\u00d72\u00d78.85\u00d710\u221212\u00d76.62\u00d710\u221234
\n\u21d2\u03bd3=7.27\u00d7105m\/s\u21d2\u03bd3=7.27\u00d7105m\/s
\nClearly, the speeds of the electron in a hydrogen atom in the levels n=1,2n=1,2 and 33 are 2.18\u00d7106m\/s2.18\u00d7106m\/s , 1.09\u00d7106m\/s1.09\u00d7106m\/s and 7.27\u00d7105m\/s7.27\u00d7105m\/s respectively.<\/h3>\n

2. Calculate the orbital period in each of these levels.<\/h2>\n

Ans: Consider T1T1 to be the orbital period of the electron when it is in level n1=1n1=1 .
\nIt is known that the orbital period is related to the orbital speed as
\nT1=2\u03c0r1\u03bd1T1=2\u03c0r1\u03bd1
\nHere,
\nr1=r1= Radius of the orbit in n1n1=n21h2\u22080\u03c0me2=n12h2\u22080\u03c0me2
\nh=h= Planck\u2019s constant =6.6\u00d710\u221234Js=6.6\u00d710\u221234Js
\ne=e= Charge of an electron =1.6\u00d710\u221219C=1.6\u00d710\u221219C
\n\u22080=\u22080= Permittivity of free space =8.85\u00d710\u221212N\u22121C2m\u22122=8.85\u00d710\u221212N\u22121C2m\u22122
\nm=m= Mass of an electron =9.1\u00d710\u221231kg=9.1\u00d710\u221231kg
\n\u21d2T1=2\u03c0\u00d7(1)2\u00d7(6.62\u00d710\u221234)2\u00d78.85\u00d710\u2212122.18\u00d7106\u00d7\u03c0\u00d79.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2\u21d2T1=2\u03c0\u00d7(1)2\u00d7(6.62\u00d710\u221234)2\u00d78.85\u00d710\u2212122.18\u00d7106\u00d7\u03c0\u00d79.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2
\n\u21d2T1=15.27\u00d710\u221217\u21d2T1=15.27\u00d710\u221217
\n\u21d2T1=1.527\u00d710\u221216s\u21d2T1=1.527\u00d710\u221216s
\nFor level n2=2n2=2, we can write the orbital period as;
\nT2=2\u03c0r2\u03bd2T2=2\u03c0r2\u03bd2
\nHere,
\nr2=r2= Radius of the orbit in n2n2=n22h2\u22080\u03c0me2=n22h2\u22080\u03c0me2
\n\u21d2T1=2\u03c0\u00d7(2)2\u00d7(6.62\u00d710\u221234)2\u00d78.85\u00d710\u2212121.09\u00d7106\u00d7\u03c0\u00d79.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2=1.22\u00d710\u221215s\u21d2T1=2\u03c0\u00d7(2)2\u00d7(6.62\u00d710\u221234)2\u00d78.85\u00d710\u2212121.09\u00d7106\u00d7\u03c0\u00d79.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2=1.22\u00d710\u221215s
\nAnd for the level n3=3n3=3, we can write the orbital period as;
\nT3=2\u03c0r3\u03bd3T3=2\u03c0r3\u03bd3
\nHere,
\nr3=r3= Radius of the orbit in n3n3=n23h2\u22080\u03c0me2=n32h2\u22080\u03c0me2
\n\u21d2T3=2\u03c0\u00d7(3)2\u00d7(6.62\u00d710\u221234)2\u00d78.85\u00d710\u2212127.27\u00d7105\u00d7\u03c0\u00d79.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2=4.12\u00d710\u221215s\u21d2T3=2\u03c0\u00d7(3)2\u00d7(6.62\u00d710\u221234)2\u00d78.85\u00d710\u2212127.27\u00d7105\u00d7\u03c0\u00d79.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2=4.12\u00d710\u221215s
\nHence, the orbital periods in the levels n=1,2n=1,2 and 33 are 1.527\u00d710\u221216s1.527\u00d710\u221216s , 1.22\u00d710\u221215s1.22\u00d710\u221215s and 4.12\u00d710\u221215s4.12\u00d710\u221215s respectively.<\/h3>\n

8. The innermost electron orbit of a hydrogen atom has a radius of 5.3\u00d710\u221211m5.3\u00d710\u221211m. What are the radii of the n=2n=2 and n=3n=3 orbits?<\/h2>\n

Ans: Provided that the innermost radius, r1=5.3\u00d710\u221211mr1=5.3\u00d710\u221211m .
\nLet r2r2 be the radius of the orbit at n=2n=2 . It is related to the radius of the innermost orbit as;
\nr2=(n)2r1r2=(n)2r1
\n\u21d2r2=(2)2\u00d75.3\u00d710\u221211=2.1\u00d710\u221210m\u21d2r2=(2)2\u00d75.3\u00d710\u221211=2.1\u00d710\u221210m
\nSimilarly, for n=3n=3;
\nr3=(n)2r1r3=(n)2r1
\n\u21d2r3=(3)2\u00d75.3\u00d710\u221211=4.77\u00d710\u221210m\u21d2r3=(3)2\u00d75.3\u00d710\u221211=4.77\u00d710\u221210m
\nClearly, the radii of the n=2n=2 and n=3n=3 orbits are 2.1\u00d710\u221210m2.1\u00d710\u221210m and 4.77\u00d710\u221210m4.77\u00d710\u221210m respectively.<\/h3>\n

9. A 12.5eV12.5eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?<\/h2>\n

Ans: It is provided that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5eV12.5eV .
\nIt is also known that the energy of the gaseous hydrogen in its ground state at room temperature is \u221213.6eV\u221213.6eV .
\nWhen gaseous hydrogen is bombarded with an electron beam at room temperature, the energy of the gaseous hydrogen becomes
\n\u221213.6+12.5eV=\u22121.1eV\u221213.6+12.5eV=\u22121.1eV
\n.
\nNow, the orbital energy is related to orbit level
\n(n)(n)
\nas;
\nE=\u221213.6(n)2eVE=\u221213.6(n)2eV
\nFor n=3;E=\u221213.69=\u22121.5eVn=3;E=\u221213.69=\u22121.5eV
\nThis energy is approximately equal to the energy of gaseous hydrogen.
\nSo, it can be concluded that the electron has excited from
\nn=1ton=3n=1ton=3
\nlevel.
\nDuring its de-excitation, the electrons can jump from
\nn=3ton=1n=3ton=1
\ndirectly, which forms a line of the Lyman series of the hydrogen spectrum.
\nThe formula for wave number for Lyman series is given as;
\n1\u03bb=Ry(112\u22121n2)1\u03bb=Ry(112\u22121n2)
\nHere,
\nRy=Ry= Rydberg constant =1.097\u00d7107m\u22121=1.097\u00d7107m\u22121
\n\u03bb=\u03bb=
\nWavelength of radiation emitted by the transition of the electron
\nUsing this relation for n=3n=3 we get,
\n1\u03bb=1.097\u00d7107(112\u2212132)1\u03bb=1.097\u00d7107(112\u2212132)
\n\u21d21\u03bb=1.097\u00d7107(1\u221219)\u21d21\u03bb=1.097\u00d7107(1\u221219)
\n\u21d21\u03bb=1.097\u00d7107(89)\u21d21\u03bb=1.097\u00d7107(89)
\n\u21d2\u03bb=98\u00d71.097\u00d7107=102.55nm\u21d2\u03bb=98\u00d71.097\u00d7107=102.55nm
\nIf the transition takes place from n=3ton=2n=3ton=2 , and then from
\nn=2ton=1n=2ton=1
\n, then the wavelength of the radiation emitted in transition from
\nn=3ton=2n=3ton=2
\nis given as;
\n1\u03bb=1.097\u00d7107(122\u2212132)1\u03bb=1.097\u00d7107(122\u2212132)
\n\u21d21\u03bb=1.097\u00d7107(14\u221219)\u21d21\u03bb=1.097\u00d7107(14\u221219)
\n\u21d21\u03bb=1.097\u00d7107(536)\u21d21\u03bb=1.097\u00d7107(536)
\n\u21d2\u03bb=365\u00d71.097\u00d7107=656.33nm\u21d2\u03bb=365\u00d71.097\u00d7107=656.33nm
\nThis radiation corresponds to the Balmer series of the hydrogen spectrum.
\nNow, the wavelength of the radiation when the transition takes place from
\nn=2ton=1n=2ton=1
\nis given as;
\n1\u03bb=1.097\u00d7107(112\u2212122)1\u03bb=1.097\u00d7107(112\u2212122)
\n\u21d21\u03bb=1.097\u00d7107(1\u221214)\u21d21\u03bb=1.097\u00d7107(1\u221214)
\n\u21d21\u03bb=1.097\u00d7107(34)\u21d21\u03bb=1.097\u00d7107(34)
\n\u21d2\u03bb=43\u00d71.097\u00d7107=121.54nm\u21d2\u03bb=43\u00d71.097\u00d7107=121.54nm
\nClearly, in the Lyman series, two wavelengths are emitted i.e., 102.5nm and 121.5nm whereas in the Balmer series, only one wavelength is emitted i.e., 656.33nm.<\/h3>\n

10. In accordance with the Bohr\u2019s model, what is the quantum number that characterizes the earth\u2019s revolution around the sun in an orbit of radius 1.5\u00d71011m1.5\u00d71011m with an orbital speed of 3\u00d7104m\/s3\u00d7104m\/s . The mass of the earth is given as 6\u00d71024kg6\u00d71024kg .<\/h2>\n

Ans: Here, it is provided that,
\nRadius of the earth\u2019s orbit around the sun, r=1.5\u00d71011mr=1.5\u00d71011m
\nOrbital speed of the earth, \u03bd=3\u00d7104m\/s\u03bd=3\u00d7104m\/s
\nMass of the earth, m=6\u00d71024kgm=6\u00d71024kg
\nWith respect to the Bohr\u2019s model, angular momentum is quantized and is given as;
\nm\u03bdr=nh2\u03c0m\u03bdr=nh2\u03c0
\nHere,
\nh=h= Planck\u2019s constant =6.6\u00d710\u221234Js=6.6\u00d710\u221234Js
\nn=n= Quantum number
\nn=m\u03bdr2\u03c0hn=m\u03bdr2\u03c0h
\n\u21d2n=2\u03c0\u00d76\u00d71024\u00d73\u00d7104\u00d71.5\u00d710116.62\u00d710\u221234\u21d2n=2\u03c0\u00d76\u00d71024\u00d73\u00d7104\u00d71.5\u00d710116.62\u00d710\u221234
\n\u21d2n=25.61\u00d71073=2.6\u00d71074\u21d2n=25.61\u00d71073=2.6\u00d71074
\nClearly, the quantum number that characterizes the earth\u2019s revolution around the sun is 2.6\u00d710742.6\u00d71074 .<\/h3>\n

11. Which of the following questions help you understand the difference between Thomson’s model and Rutherford’s model better.
\n1. Is the average angle of deflection of \u03b1\u2212\u03b1\u2212particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?<\/h2>\n

Ans: About the same.
\nThe average angle of deflection of \u03b1\u2212\u03b1\u2212particles caused by a thin gold foil considered by Thomson’s model is about the same size as that considered by Rutherford’s model.
\nThis is because in both the models, the average angle was used.<\/h3>\n

2. Is the probability of backward scattering (i.e., scattering of \u03b1\u2212\u03b1\u2212particles at angles greater than 90\u221890\u2218) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?<\/h2>\n

Ans: Much less.
\nThe probability of scattering of \u03b1\u2212\u03b1\u2212particles at angles greater than 90\u00b0 considered by Thomson’s model is much less than that considered by Rutherford’s model.<\/h3>\n

3. Keeping other factors fixed, it is found experimentally that for small thickness t, the number of \u03b1\u2212\u03b1\u2212particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?<\/h2>\n

Ans: Scattering is mainly caused due to single collisions.
\nThe chances of a single collision have a linearly increasing nature with the number of target atoms.
\nAs the number of target atoms increases with an increase in thickness, the collision probability varies linearly with the thickness of the target.<\/h3>\n

4. In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of \u03b1\u2212\u03b1\u2212particles by a thin foil?<\/h2>\n

Ans: Thomson’s model.
\nIt is incorrect to not consider multiple scattering in Thomson’s model for the calculation of average angle of scattering of \u03b1\u2212\u03b1\u2212particles by a thin foil.
\nThis is because a single collision produces very little deflection in this model.
\nThus, the observed average scattering angle can be demonstrated only by considering multiple scattering.<\/h3>\n

12. The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10\u22124010\u221240. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.<\/h2>\n

Ans: The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10\u22124010\u221240.
\nIt is given that,
\nRadius of the first Bohr orbit is given by the relation,
\nr1=4\u03c0\u22080(h2\u03c0)2mee2r1=4\u03c0\u22080(h2\u03c0)2mee2 \u2026(1)
\nHere,
\n\u22080=\u22080= Permittivity of free space
\nh=h= Planck\u2019s constant =6.63\u00d710\u221234Js=6.63\u00d710\u221234Js
\nme=me= Mass of an electron =9.1\u00d710\u221231kg=9.1\u00d710\u221231kg
\ne=e= Charge of an electron =1.6\u00d710\u221219C=1.6\u00d710\u221219C
\nmp=mp= Mass of a proton =1.67\u00d710\u221227kg=1.67\u00d710\u221227kg
\nr=r= Distance between the electron and the proton
\nCoulomb attraction between an electron and a proton is given by;
\nFC=e24\u03c0\u22080r2FC=e24\u03c0\u22080r2 \u2026(2)
\nGravitational force of attraction between an electron and a proton is given by;
\nFG=Gmpmer2FG=Gmpmer2 \u2026(3)
\nHere,
\nG=G= Gravitational constant =6.67\u00d710\u221211Nm2\/kg2=6.67\u00d710\u221211Nm2\/kg2
\nThe electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write;
\nFG=FCFG=FC
\n\u21d2Gmpmer2=e24\u03c0\u22080r2\u21d2Gmpmer2=e24\u03c0\u22080r2 \u2026(using 2 & 3)
\n\u21d2e24\u03c0\u22080=Gmpme\u21d2e24\u03c0\u22080=Gmpme \u2026(4)
\nNow, by substituting the value of equation (4) in (1), we get;
\nr1=(h2\u03c0)2Gmpmer1=(h2\u03c0)2Gmpme
\n\u21d2r1=(6.63\u00d710\u2212342\u00d73.14)26.67\u00d710\u221211\u00d71.67\u00d710\u221227\u00d7(9.1\u00d710\u221231)2\u21d2r1=(6.63\u00d710\u2212342\u00d73.14)26.67\u00d710\u221211\u00d71.67\u00d710\u221227\u00d7(9.1\u00d710\u221231)2
\n\u21d2r1\u22481.21\u00d71029m\u21d2r1\u22481.21\u00d71029m
\nIt is known that the universe is 156 billion light years wide or 1.5\u00d71027m1.5\u00d71027m wide. Clearly, it can be concluded that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.<\/h3>\n

13. Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level nn to level (n\u22121)(n\u22121). For large nn , show that this frequency equals the classical frequency of revolution of the electron in the orbit.<\/h2>\n

Ans: It is given that a hydrogen atom makes transition from an upper level (n)(n) to a lower level (n\u22121)(n\u22121). We have the relation for energy (E1)(E1) of radiation at level nn as;
\nE1=h\u03bd1=hme4(4\u03c0)3\u220820(h2\u03c0)3\u00d7(1n2)E1=h\u03bd1=hme4(4\u03c0)3\u220802(h2\u03c0)3\u00d7(1n2) \u2026(1)
\nHere,
\n\u03bd1=\u03bd1= Frequency of radiation at level nn
\nh=h= Planck\u2019s constant
\nm=m= Mass of hydrogen atom
\ne=e= Charge on an electron
\n\u22080=\u22080= Permittivity of free space
\nNow, the relation for energy (E2)(E2) of radiation at level (n\u22121)(n\u22121) is given as;
\nE2=h\u03bd2=hme4(4\u03c0)3\u220820(h2\u03c0)3\u00d71(n\u22121)2E2=h\u03bd2=hme4(4\u03c0)3\u220802(h2\u03c0)3\u00d71(n\u22121)2 \u2026(2)
\nHere,
\n\u03bd2=\u03bd2= Frequency of radiation at level (n\u22121)(n\u22121)
\nEnergy (E)(E) released as a result of de-excitation;
\nE=E2\u2212E1E=E2\u2212E1
\n\u21d2h\u03bd=E2\u2212E1\u21d2h\u03bd=E2\u2212E1 \u2026(3)
\nHere,
\n\u03bd=\u03bd= Frequency of radiation emitted
\nUsing the values of equations (1) and (2) in equation (3), we get;
\n\u03bd=me4(4\u03c0)3\u220820(h2\u03c0)3[1(n\u22121)2\u22121n2]\u03bd=me4(4\u03c0)3\u220802(h2\u03c0)3[1(n\u22121)2\u22121n2]
\n\u21d2\u03bd=me4(2n\u22121)(4\u03c0)3\u220820(h2\u03c0)3n2(n\u22121)2\u21d2\u03bd=me4(2n\u22121)(4\u03c0)3\u220802(h2\u03c0)3n2(n\u22121)2
\nFor large nn, we can write (2n\u22121)\u22122n(2n\u22121)\u22122n and (n\u22121)\u2243n(n\u22121)\u2243n
\n\u21d2\u03bd=me432\u03c03\u220820(h2\u03c0)3n3\u21d2\u03bd=me432\u03c03\u220802(h2\u03c0)3n3 \u2026(4)
\nClassical relation of frequency of revolution of an electron is given by;
\n\u03bdc=\u03bd2\u03c0r\u03bdc=\u03bd2\u03c0r \u2026(5)
\nHere,
\nVelocity of the electron in the nthnth orbit is given as;
\n\u03bd=e24\u03c0\u22080(h2\u03c0)n\u03bd=e24\u03c0\u22080(h2\u03c0)n \u2026(6)
\nAnd, radius of the nthnth orbit is given by;
\nr=4\u03c0\u22080(h2\u03c0)2me2n2r=4\u03c0\u22080(h2\u03c0)2me2n2 \u2026(7)
\nSubstituting the values of equation (6) and (7) in equation (5), we get;
\n\u03bdc=me432\u03c03\u220820(h2\u03c0)3n3\u03bdc=me432\u03c03\u220802(h2\u03c0)3n3 \u2026(8)
\nClearly, when equations (4) and (8) are compared, it can be seen that the frequency of radiation emitted by the hydrogen atom is the same as the classical orbital frequency.<\/h3>\n

14. Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (\u223c10\u221210m)(\u223c10\u221210m).
\n1. Construct a quantity with the dimensions of length from the fundamental constants e,meandce,meandc. Also, determine its numerical value.<\/h2>\n

Ans: To construct a quantity with the dimensions of length from the fundamental constants e,meandce,meandc, take:
\nCharge of an electron, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C
\nMass of an electron, me=9.1\u00d710\u221231kgme=9.1\u00d710\u221231kg
\nSpeed of light, c=3\u00d7108m\/sc=3\u00d7108m\/s
\nThe quantity having dimensions of length and involving the given quantities is (e24\u03c0\u22080mec2)(e24\u03c0\u22080mec2)
\nHere,<\/h3>\n

\u22080=\u22080= Permittivity of free space
\nAnd, 14\u03c0\u22080=9\u00d7109Nm2C\u2212214\u03c0\u22080=9\u00d7109Nm2C\u22122
\n\u2234\u2234The numerical value of the taken quantity will be
\n14\u03c0\u22080\u00d7e2mec2=9\u00d7109\u00d7(1.6\u00d710\u221219)29.1\u00d710\u221231\u00d7(3\u00d7108)2=2.81\u00d710\u221215m14\u03c0\u22080\u00d7e2mec2=9\u00d7109\u00d7(1.6\u00d710\u221219)29.1\u00d710\u221231\u00d7(3\u00d7108)2=2.81\u00d710\u221215m
\nClearly, the numerical value of the taken quantity is much less than the typical size of an atom.<\/h3>\n

2. You will observe that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for \u201csomething else\u201d to get the right atomic size. Now, the Planck’s constant hh had already made its appearance elsewhere. Bohr\u2019s great insight lay in recognising that h,\u22b8me,\u22b8and\u22b8eh,\u22b8me,\u22b8and\u22b8e
\nwill yield the right atomic size. Construct a quantity with the dimension of length from h,meandeh,meande
\nand confirm that its numerical value has indeed the correct order of magnitude.<\/h2>\n

Ans: To construct a quantity with the dimension of length from
\nh,meandeh,meande
\n, take,
\nCharge of an electron, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C
\nMass of an electron, me=9.1\u00d710\u221231kgme=9.1\u00d710\u221231kg
\nPlanck\u2019s constant, h=6.63\u00d710\u221234Jsh=6.63\u00d710\u221234Js
\nNow, let us take a quantity involving the given quantities as, 4\u03c0\u22080(h22\u03c0)mee24\u03c0\u22080(h22\u03c0)mee2
\nHere, \u22080=\u22080= Permittivity of free space
\nAnd, 14\u03c0\u22080=9\u00d7109Nm2C\u2212214\u03c0\u22080=9\u00d7109Nm2C\u22122
\n\u2234\u2234The numerical value of the taken quantity would be
\n4\u03c0\u22080\u00d7(h2\u03c0)2mee2=19\u00d7109\u00d7(6.63\u00d710\u2212342\u00d73.14)29.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2=0.53\u00d710\u221210m4\u03c0\u22080\u00d7(h2\u03c0)2mee2=19\u00d7109\u00d7(6.63\u00d710\u2212342\u00d73.14)29.1\u00d710\u221231\u00d7(1.6\u00d710\u221219)2=0.53\u00d710\u221210m
\nClearly, the value of the quantity taken is of the order of the atomic size.<\/h3>\n

15. The total energy of an electron in the first excited state of the hydrogen atom is about \u22123.4eV\u22123.4eV.
\n1. What is the kinetic energy of the electron in this state?<\/h2>\n

Ans: Kinetic energy of the electron is the same as the negative of the total energy.
\n\u21d2K=\u2212E\u21d2K=\u2212E
\n\u21d2K=\u2212(\u22123.4)=+3.4eV\u21d2K=\u2212(\u22123.4)=+3.4eV
\nClearly, the kinetic energy of the electron in the given state is +3.4eV+3.4eV.<\/h3>\n

2. What is the potential energy of the electron in this state?<\/h2>\n

Ans: Potential energy (U)(U) of the electron is the same as the negative of twice of its kinetic energy
\n\u21d2U=\u22122K\u21d2U=\u22122K
\n\u21d2U=\u2212(\u22123.4)=\u22126.8eV\u21d2U=\u2212(\u22123.4)=\u22126.8eV
\nClearly, the potential energy of the electron in the given state is \u22126.8eV\u22126.8eV.<\/h3>\n

3. Which of the answers above would change if the choice of the zero of potential energy is changed?<\/h2>\n

Ans: The potential energy of a system is dependent on the reference point taken. Here, the potential energy of the reference point is considered to be zero.
\nWhen the reference point is changed, then the magnitude of the potential energy of the system also changes.
\nAs total energy is the sum of kinetic and potential energies, total energy of the system would also differ.<\/h3>\n

16. If Bohr’s quantization postulate (angular momentum =nh\/2n=nh\/2n) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun?<\/h2>\n

Ans: It is not much spoken about the quantization of orbits of planets around the Sun since the angular momentum associated with planetary motion is largely relative to the value of constant (h)(h).
\nThe angular momentum of the Earth in its orbit is of the order of 1070h1070h. This causes a very high value of quantum levels nn of the order of 10701070.
\nWhen large values of nn are considered, successive energies and angular momenta are found to be relatively very small. Clearly, the quantum levels for planetary motion are considered continuous.<\/h3>\n

17. Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom (i.e., an atom in which a negatively charged muon (\u03bc\u2212)(\u03bc\u2212)mass about 207me207me. orbits around a proton).<\/h2>\n

Ans: To obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom, consider the mass of a negat ively charged muon to be m\u03bc=207mm\u03bc=207m.
\nNow, according to Bohr\u2019s model,
\nBohr radius, re\u221d(1me)re\u221d(1me)
\nAnd, energy of a ground state electronic hydrogen atom, Ee\u221dmeEe\u221dme.
\nAlso, energy of a ground state muonic hydrogen atom, E\u03bc\u221dmeE\u03bc\u221dme.
\nIt is known that the value of the first Bohr orbit, re=0.53A=0.53\u00d710\u221210mre=0.53A=0.53\u00d710\u221210m
\nConsider r\u03bcr\u03bc to be the radius of muonic hydrogen atom.
\nAt equilibrium, we have the relation:
\nm\u03bcr\u03bc=merem\u03bcr\u03bc=mere
\n\u21d2207me\u00d7r\u03bc=mere\u21d2207me\u00d7r\u03bc=mere
\n\u21d2r\u03bc=0.53\u00d710\u221210207=2.56\u00d710\u221213m\u21d2r\u03bc=0.53\u00d710\u221210207=2.56\u00d710\u221213m
\nClearly, the value of the first Bohr radius of muonic hydrogen atom is 2.56\u00d710\u221213m2.56\u00d710\u221213m.
\nNow, we have,
\nEe=\u221213.6eVEe=\u221213.6eV
\nTaking the ratio of these energies as EeE\u03bc=mem\u03bcEeE\u03bc=mem\u03bc
\n\u21d2EeE\u03bc=me207me\u21d2EeE\u03bc=me207me
\n\u21d2E\u03bc=207Ee\u21d2E\u03bc=207Ee
\n\u21d2E\u03bc=207\u00d7(\u221213.6)=\u22122.81keV\u21d2E\u03bc=207\u00d7(\u221213.6)=\u22122.81keV
\nClearly, the ground state energy of a muonic hydrogen atom is \u22122.81keV\u22122.81keV.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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