{"id":120436,"date":"2022-05-10T12:38:13","date_gmt":"2022-05-10T07:08:13","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=120436"},"modified":"2022-05-10T12:38:13","modified_gmt":"2022-05-10T07:08:13","slug":"chapter-8-electromagnetic-waves-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-8-electromagnetic-waves-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 8 \u2013 Electromagnetic Waves Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"

1. Figure Drawn below Shows a Capacitor Made of Two Circular Plates Each of Radius
\n12 cm12 cm
\n, and Separated by
\n5.0 cm5.0 cm
\n. The capacitor is being Charged by an External Source (not Shown in the Figure). The Charging Current is Constant and Equal to
\n0.15 A0.15 A<\/h2>\n

1. Calculate the Capacitance and the Rate of Change of the Potential Difference Between the Plates.<\/h2>\n

Ans: Radius of each circular plates,
\nr=12 cm=0.12 mr=12 cm=0.12 m
\nDistance between the given plates,
\nd=5 cm=0.05 md=5 cm=0.05 m
\nCharging current, I=0.15AI=0.15A
\nThe permittivity of free space, \u03b50=8.85\u00d710\u221212C2N\u22121m\u22122\u03b50=8.85\u00d710\u221212C2N\u22121m\u22122
\nCapacitance between the two plates can be given as:
\nC=\u03b50AdC=\u03b50Ad
\nWhere,
\nA=\u03c0r2A=\u03c0r2.
\nHence,
\nC=\u03b50\u03c0r2dC=\u03b50\u03c0r2d
\nC=8.85\u00d710\u221212\u00d7(0.12)20.05C=8.85\u00d710\u221212\u00d7(0.12)20.05
\nC=8.0032\u00d710\u221212FC=8.0032\u00d710\u221212F
\nC=80.032pFC=80.032pF
\nCharge on each plate is given as,
\nq=CVq=CV
\n, where,
\nV = Potential difference across the plates
\nDifferentiating both sides with respect to time (t), we get:
\ndqdt=CdVdtdqdt=CdVdt
\nAs, dqdt=Idqdt=I, so, the rate of change of potential difference between the plates can be given as:
\ndVdt=IC=0.1580.032\u00d710\u221212dVdt=IC=0.1580.032\u00d710\u221212
\n\u2234dVdt=1.87\u00d7109V\/s\u2234dVdt=1.87\u00d7109V\/s<\/h3>\n

2. Obtain the Displacement Current Across the Plates.<\/h2>\n

Ans: The displacement current across the plates would be the same as the conduction current. Hence, the displacement current, IdId would be
\n0.15 A0.15 A
\n.<\/h3>\n

3. Is Kirchhoff\u2019s First Rule (junction Rule) Valid at Each Plate of the Capacitor? Explain.<\/h2>\n

Ans: Yes, Kirchhoff\u2019s first rule would be valid at each plate of the capacitor, provided that sum of conduction and displacement currents, i.e., I=Ic+IdI=Ic+Id (junction rule of Kirchhoff\u2019s law).<\/h3>\n

2) A Parallel Plate Capacitor (figure) Made of Circular Plates Each of Radius
\nR=6.0 cmR=6.0 cm
\nhas a capacitance
\nC=100 pFC=100 pF
\n. The Capacitor is Connected to a
\n230 V230 V
\nac supply with a (angular) frequency of
\n300rads\u22121300rads\u22121<\/h3>\n

1. What is the RMS Value of the Conduction Current?<\/h2>\n

Ans: Radius of each circular plate,
\nR=6.0 cm=0.06 mR=6.0 cm=0.06 m
\nCapacitance of a parallel plate capacitor,
\nC=100 pF=100\u00d710\u221212FC=100 pF=100\u00d710\u221212F
\nSupply voltage,
\nV=230 VV=230 V
\nAngular frequency, \u03c9=300rads\u22121\u03c9=300rads\u22121
\nRms value of conduction current can be given as: I=VXCI=VXC.
\nWhere, XCXC = capacitive currant =1\u03c9C=1\u03c9C
\n\u2234I=V\u00d7\u03c9C\u2234I=V\u00d7\u03c9C
\nI=230\u00d7300\u00d7100\u00d710\u221212I=230\u00d7300\u00d7100\u00d710\u221212
\nI=6.9\u00d710\u22126I=6.9\u00d710\u22126
\nI=6.9\u03bcAI=6.9\u03bcA
\nTherefore, the RMS value of conduction current will be 6.9\u03bcA6.9\u03bcA.<\/h3>\n

2. Is the Conduction Current Equal to the Displacement Current?<\/h2>\n

Ans: Yes, the conduction current will be the same as the displacement current. i.e., the conduction current is equal to the displacement current.<\/h3>\n

3. Determine the Amplitude of B\u2192B\u2192 at a point
\n3.0 cm3.0 cm
\nfrom the Axis Between the Plates.<\/h2>\n

Ans: Magnetic field is given as: B\u2192=\u03bc0r2\u03c0R2I0B\u2192=\u03bc0r2\u03c0R2I0
\nWhere,
\n\u03bc0\u03bc0= Free space permeability =4\u03c0\u00d710\u22127NA\u22122=4\u03c0\u00d710\u22127NA\u22122
\nI0I0 = Maximum value of current = 2\u2013\u221aI2I
\nrr= Distance between the plates from the axis = 3.0cm=0.03cm3.0cm=0.03cm.
\n\u2234B\u2192=4\u03c0\u00d710\u22127\u00d70.03\u00d72\u221a\u00d76.9\u00d710\u221262\u03c0\u00d7(0.06)2\u2234B\u2192=4\u03c0\u00d710\u22127\u00d70.03\u00d72\u00d76.9\u00d710\u221262\u03c0\u00d7(0.06)2
\nB\u2192=1.63\u00d710\u221211TB\u2192=1.63\u00d710\u221211T
\nTherefore, the magnetic field at that point will be 1.63\u00d710\u221211T1.63\u00d710\u221211T.<\/h3>\n

3. What Physical Quantity is the Same for X-Rays of Wavelength
\n10\u221210m10\u221210m
\n, red light of wavelength
\n6800A\u22186800A\u2218
\nand radio waves of wavelength
\n500 m500 m
\n?<\/h2>\n

Ans: The speed of light (
\n3\u00d7108m\/s3\u00d7108m\/s
\n) is independent of the wavelength in the vacuum. Hence, it is the same for all wavelengths in a vacuum.<\/h3>\n

4. A Plane Electromagnetic Wave Travels in Vacuum Along Z-Direction. What Can You Say About the Directions of Its Electric and Magnetic Field Vectors? If the Frequency of the Wave is
\n30MHz30MHz
\n, what is its wavelength?<\/h2>\n

Ans: The electromagnetic wave is travelling along the z-direction, in a vacuum. The electric field (E) and the magnetic field (H) will lie in the x-y plane and they will be mutually perpendicular.
\nFrequency of the wave,
\n\u03bd=30 Mhz=30\u00d7106s\u22121\u03bd=30 MHz=30\u00d7106s\u22121
\nSpeed of light in a vacuum,
\nc=3\u00d7108m\/sc=3\u00d7108m\/s
\nWavelength, \u03bb\u03bb of the wave can be given as: \u03bb=c\u03bd\u03bb=c\u03bd
\n\u2234\u03bb=3\u00d710830\u00d7106=10m\u2234\u03bb=3\u00d710830\u00d7106=10m<\/h3>\n

5. A Radio Can Tune in to Any Station in the
\n7.5 Mhz7.5 MHz
\nto
\n12 Mhz12 MHz
\nBand. What is the corresponding Wavelength Band?<\/h2>\n

Ans: A radio can tune to minimum frequency,
\n\u03bd1=7.5 Mhz=7.5\u00d7106Hz\u03bd1=7.5 MHz=7.5\u00d7106Hz
\nMaximum frequency,
\n\u03bd2=12 Mhz=12\u00d7106Hz\u03bd2=12 MHz=12\u00d7106Hz
\nSpeed of light,
\nc=3\u00d7108m\/sc=3\u00d7108m\/s
\nWavelength for frequency, \u03bd1\u03bd1 can be calculated as:
\n\u03bb1=c\u03bd1\u03bb1=c\u03bd1
\n\u03bb1=3\u00d71087.5\u00d7106\u03bb1=3\u00d71087.5\u00d7106
\n\u03bb1=40m\u03bb1=40m
\nWavelength for frequency, \u03bd2\u03bd2 can be calculated as:
\n\u03bb2=c\u03bd2\u03bb2=c\u03bd2
\n\u03bb2=3\u00d710812\u00d7106\u03bb2=3\u00d710812\u00d7106
\n\u03bb2=25m\u03bb2=25m
\nTherefore, the corresponding wavelength band would be between 40m40m to 25m25m.<\/h3>\n

6. A Charged Particle Oscillates About Its Mean Equilibrium Position With a Frequency of 109Hz109Hz. What is the Frequency of the Electromagnetic Waves Produced by the Oscillator?<\/h2>\n

Ans: The frequency of an electromagnetic wave produced by the oscillator will be equal to the frequency of charged particle oscillating about its mean position i.e.,
\n109Hz109Hz
\n.<\/h3>\n

7. The Amplitude of the Magnetic Field Part of a Harmonic Electromagnetic Wave in Vacuum is B0 = 510nT. What is the Amplitude of the Electric Field Part of the Wave?<\/h2>\n

Ans: Amplitude of magnetic field of an electromagnetic wave in a vacuum, is:
\nB0=510nT=510\u00d710\u22129TB0=510nT=510\u00d710\u22129T
\nSpeed of light in a vacuum, c=3\u00d7108m\/sc=3\u00d7108m\/s
\nAmplitude of electric field of the electromagnetic wave can be given as:
\nE=cB=3\u00d7108\u00d7510\u00d710\u22129=153N\/CE=cB=3\u00d7108\u00d7510\u00d710\u22129=153N\/C.<\/h3>\n

8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120N\/C and that its frequency is
\n\u03bd=50.0 Mhz\u03bd=50.0 MHz
\n.
\n1. Determine, B0B0, \u03c9\u03c9, kk, and \u03bb\u03bb.<\/h2>\n

Ans: Electric field amplitude, E0=120N\/CE0=120N\/C
\nFrequency of source,
\n\u03bd=50.0MHz=50\u00d7106Hz\u03bd=50.0MHz=50\u00d7106Hz
\nSpeed of light,
\nc=3\u00d7108m\/sc=3\u00d7108m\/s
\nMagnetic field strength can be given as:
\nB0=E0cB0=E0c
\nB0=1203\u00d7108B0=1203\u00d7108
\nB0=4\u00d710\u22127TB0=4\u00d710\u22127T
\nB0=400nTB0=400nT
\nAngular Frequency of Source Can be Given As:
\n\u03c9=2\u03c0\u03bd\u03c9=2\u03c0\u03bd
\n\u03c9=2\u03c0(50\u00d7106)\u03c9=2\u03c0(50\u00d7106)
\n\u03c9=3.14\u00d7108rad\/s\u03c9=3.14\u00d7108rad\/s
\nPropagation Constant Can be Given As:
\nk=\u03c9ck=\u03c9c
\nk=3.14\u00d71083\u00d7108k=3.14\u00d71083\u00d7108
\nk=1.05rad\/mk=1.05rad\/m
\nWavelength of Wave Can be Given As:
\n\u03bb=c\u03bd\u03bb=c\u03bd
\n\u03bb=3\u00d710850\u00d7106\u03bb=3\u00d710850\u00d7106
\n\u03bb=6.0m\u03bb=6.0m<\/h3>\n

2. Find Expressions for E and B.<\/h2>\n

Ans: If the wave propagates in the positive x-direction. Then, the electric field vector would be in the positive y-direction and the magnetic field vector will lie in the positive z-direction. This is because all three vectors are mutually perpendicular.
\nEquation of electric field vector can be given as:
\nE\u2192=E0sin(kx\u2212\u03c9t)j^E\u2192=E0sin(kx\u2212\u03c9t)j^
\nE\u2192=120sin[1.05x\u22123.14\u00d7108t]j^E\u2192=120sin\u2061[1.05x\u22123.14\u00d7108t]j^
\nAnd, magnetic field vector can be given as:
\nB\u2192=B0sin(kx\u2212\u03c9t)k^B\u2192=B0sin(kx\u2212\u03c9t)k^
\nB\u2192=(4\u00d710\u22127)sin[1.05x\u22123.14\u00d7108t]k^B\u2192=(4\u00d710\u22127)sin\u2061[1.05x\u22123.14\u00d7108t]k^<\/h3>\n

9. The Terminology of Different Parts of the Electromagnetic Spectrum Is Given in the Text. Use the Formula E = hv (for Energy of a Quantum of Radiation: Photon) and Obtain the Photon Energy in Units of ev for Different Parts of the Electromagnetic Spectrum. in What Way are the Different Scales of Photon Energies That You Obtain Related to the Sources of Electromagnetic Radiation?<\/h2>\n

Ans: The energy of photon can be given as:
\nE=h\u03bd=hc\u03bbE=h\u03bd=hc\u03bb
\nWhere,
\nhh = Planck\u2019s constant =6.6\u00d710\u221234Js=6.6\u00d710\u221234Js
\ncc = speed of light =3\u00d7108m\/s=3\u00d7108m\/s
\n\u03bb\u03bb = wavelength of radiation
\n\u2234E=6.6\u00d710\u221234\u00d73\u00d7108\u03bbJ\u2234E=6.6\u00d710\u221234\u00d73\u00d7108\u03bbJ
\nE=19.8\u00d710\u221226\u03bbJE=19.8\u00d710\u221226\u03bbJ
\nE=19.8\u00d710\u221226\u03bb\u00d71.6\u00d710\u221219eVE=19.8\u00d710\u221226\u03bb\u00d71.6\u00d710\u221219eV
\nE=12.375\u00d710\u22127\u03bbeVE=12.375\u00d710\u22127\u03bbeV
\nFor Different Values of in an Electromagnetic Spectrum, Photon Energies
\nare Listed in Below Table:<\/h3>\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n\n\n\n
\u03bb(m)\u03bb(m)<\/td>\nE(eV)E(eV)<\/td>\n<\/tr>\n
103103<\/td>\n12.375\u00d710\u22121012.375\u00d710\u221210<\/td>\n<\/tr>\n
1<\/td>\n12.375\u00d710\u2212712.375\u00d710\u22127<\/td>\n<\/tr>\n
10\u2212310\u22123<\/td>\n12.375\u00d710\u2212412.375\u00d710\u22124<\/td>\n<\/tr>\n
10\u2212610\u22126<\/td>\n12.375\u00d710\u2212112.375\u00d710\u22121<\/td>\n<\/tr>\n
10\u2212810\u22128<\/td>\n12.375\u00d710112.375\u00d7101<\/td>\n<\/tr>\n
10\u22121010\u221210<\/td>\n12.375\u00d710312.375\u00d7103<\/td>\n<\/tr>\n
10\u22121210\u221212<\/td>\n12.375\u00d710512.375\u00d7105<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\n10. In a Plane Electromagnetic Wave, the Electric Field Oscillates Sinusoidally at a Frequency of 2\u00d71010Hz2\u00d71010Hz and amplitude 48Vm- 1
\n1. What is the Wavelength of the Wave?<\/h2>\n

Ans: Frequency of the electromagnetic wave,
\n\u03bd=2.0\u00d71010Hz\u03bd=2.0\u00d71010Hz
\nElectric field amplitude,
\nE0=48 Vm\u22121E0=48 Vm\u22121
\nSpeed of light,
\nc=3\u00d7108m\/sc=3\u00d7108m\/s
\nWavelength of wave can be given as:
\n\u03bb=c\u03bd\u03bb=c\u03bd
\n\u03bb=3\u00d71082\u00d71010\u03bb=3\u00d71082\u00d71010
\n\u03bb=0.015m\u03bb=0.015m<\/h3>\n

2. What is the Amplitude of the Oscillating Magnetic Field?<\/h2>\n

Ans: Magnetic field strength can be given as:
\nB0=E0cB0=E0c
\n=483\u00d7108=483\u00d7108
\n=1.6\u00d710\u22127T=1.6\u00d710\u22127T<\/h3>\n

3. Show That the Average Energy Density of the E Field Equals the Average Energy Density of the B field. C = 3 x 108m\/s<\/h2>\n

Ans: Energy density of the electric field is given as:
\nUE=12\u22080E2UE=12\u22080E2
\nEnergy density of the magnetic field can be given as:
\nUB=12\u03bc0B2UB=12\u03bc0B2
\nWhere,
\n\u22080\u22080 = Permittivity of free space
\n\u03bc0\u03bc0 = Permeability of free space
\nThe relation between E and B can be given as:
\nE=cBE=cB \u2026.(1)
\nWhere,
\nc=1\u22080\u03bc0\u221ac=1\u22080\u03bc0 \u2026.(2)
\nSubstituting equation (2) in equation (1), we get:
\nE=1\u22080\u03bc0\u221aBE=1\u22080\u03bc0B
\nSquaring both the sides, we get:
\nE2=1\u22080\u03bc0B2E2=1\u22080\u03bc0B2
\nE2\u22080=B2\u03bc0E2\u22080=B2\u03bc0
\n12E2\u22080=12B2\u03bc012E2\u22080=12B2\u03bc0
\n\u21d2UE=UB\u21d2UE=UB.<\/h3>\n

11. Suppose That the Electric Field Part of an Electromagnetic Wave in Vacuum is
\nE\u2192={(3.1N\/C)cos[(1.8rad\/m)y+(5.4\u00d7106rad\/s)t]}i^E\u2192={(3.1N\/C)cos[(1.8rad\/m)y+(5.4\u00d7106rad\/s)t]}i^
\n.
\n1. What is the Direction of Propagation?<\/h2>\n

Ans: From the given electric field vector, we can say that the electric field will be directed along the negative x direction. Hence, the direction of motion will be along the negative y direction i.e., \u2212j\u2212j.<\/h3>\n

2. What is the Wavelength \u03bb\u03bb?<\/h2>\n

Ans: The general equation for electric field in the positive x direction is given as:
\nE\u2192=E0sin(kx\u2212\u03c9t)i^E\u2192=E0sin(kx\u2212\u03c9t)i^
\n. By comparing the given equation,
\nE\u2192={(3.1N\/C)cos[(1.8rad\/m)y+(5.4\u00d7106rad\/s)t]}i^E\u2192={(3.1N\/C)cos[(1.8rad\/m)y+(5.4\u00d7106rad\/s)t]}i^
\nwith general equation, we get:
\nElectric field amplitude, E0=3.1N\/CE0=3.1N\/C
\nWave number, k=1.8rad\/mk=1.8rad\/m
\nAngular frequency, \u03c9=5.4\u00d7106rad\/s\u03c9=5.4\u00d7106rad\/s
\nWavelength, \u03bb=2\u03c0k=2\u03c01.8=3.490m\u03bb=2\u03c0k=2\u03c01.8=3.490m.<\/h3>\n

3. What is the Frequency \u03bd\u03bd?<\/h2>\n

Ans: The frequency, \u03bd\u03bd of the wave can be given as: \u03bd=\u03c92\u03c0\u03bd=\u03c92\u03c0
\n=5.4\u00d71082\u03c0=8.6\u00d7107Hz=5.4\u00d71082\u03c0=8.6\u00d7107Hz.<\/h3>\n

4. What is the Amplitude of the Magnetic Field Part of the Wave?<\/h2>\n

Ans: Magnetic field strength can be given as:
\nB0=E0cB0=E0c
\n\u2234B0=3.13\u00d7108=1.03\u00d710\u22127T\u2234B0=3.13\u00d7108=1.03\u00d710\u22127T.<\/h3>\n

5. Write an Expression for the Magnetic Field Part of the Wave.<\/h2>\n

Ans: We can observe that the magnetic field vector will be directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
\nB\u2192=B0cos(ky+\u03c9t)k^B\u2192=B0cos\u2061(ky+\u03c9t)k^
\n, hence,
\nB\u2192={(1.03\u00d710\u22127T)cos[(1.8rad\/m)y+(5.4\u00d7106rad\/s)t]}k^B\u2192={(1.03\u00d710\u22127T)cos[(1.8rad\/m)y+(5.4\u00d7106rad\/s)t]}k^
\n.<\/h3>\n

12. About
\n5%5%
\nof the power of a
\n100 W100 W
\nlight bulb is converted to visible radiation. What is the average intensity of visible radiation
\n1. At a Distance of
\n1 m1 m
\nfrom the bulb? Assume That the Radiation Is Emitted Isotropically and Neglect Reflection.<\/h2>\n

Ans: Power rating of bulb,
\nP=100 WP=100 W
\nAbout
\n5%5%
\nof its power has been converted into visible radiation.
\nTherefore, power of visible radiation is given as:
\np\u2032=5100\u00d7100=5Wp\u2032=5100\u00d7100=5W
\nAverage intensity at distance, d=1md=1m can be given as:
\nI=P\u2032A=p\u20324\u03c0d2I=P\u2032A=p\u20324\u03c0d2
\nI=54\u00d73.14\u00d7(1)2=0.398W\/m2I=54\u00d73.14\u00d7(1)2=0.398W\/m2<\/h3>\n

2. At a Distance of 10m? Assume That the Radiation is Emitted Isotropically and Neglect Reflection.<\/h2>\n

Ans: Power of visible radiation can be given as:
\np\u2032=5100\u00d7100=5Wp\u2032=5100\u00d7100=5W.
\nAverage intensity at distance, d=10md=10m can be given as:
\nI=p\u2032A=p\u20324\u03c0d2I=p\u2032A=p\u20324\u03c0d2
\nI=54\u00d73.14\u00d7(10)2=0.00398W\/m2I=54\u00d73.14\u00d7(10)2=0.00398W\/m2<\/h3>\n

13. Use the Formula \u03bbm=0.29TcmK\u03bbm=0.29TcmK to Obtain the Characteristic Temperature Ranges for Different Parts of the Electromagnetic Spectrum. What Do the Numbers That You Obtain Tell You?<\/h2>\n

Ans: At a particular temperature, a body produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given by Planck\u2019s law. It can be represented as:
\n\u03bbm=0.29TcmK\u03bbm=0.29TcmK.
\nWhere,
\n\u03bbm\u03bbm = maximum wavelength
\nTT = temperature
\nTherefore, the temperature for different wavelengths can be given as:
\nFor, \u03bbm=10\u22124cm\u03bbm=10\u22124cm, T=0.2910\u22124=2900\u2218KT=0.2910\u22124=2900\u2218K
\nFor, \u03bbm=5\u00d710\u22125cm\u03bbm=5\u00d710\u22125cm, T=0.295\u00d710\u22125=5800\u2218KT=0.295\u00d710\u22125=5800\u2218K
\nFor, \u03bbm=10\u22126cm\u03bbm=10\u22126cm, T=0.2910\u22126=290000\u2218KT=0.2910\u22126=290000\u2218K
\nWe can see that, as the maximum wavelength decreases, the corresponding temperature increases. At lower temperature, wavelength produced will not have maximum intensity.<\/h3>\n

14. Given Below Are Some Famous Numbers Associated With Electromagnetic Radiations in Different Contexts in Physics. State the Part of the Electromagnetic Spectrum to Which Each Belongs.
\n1. 21 cm21 cm
\n(Wavelength Emitted by Atomic Hydrogen in Interstellar Space).<\/h2>\n

Ans: Wavelength of
\n21 cm21 cm
\nbelongs to a short radio wave,
\nwhich is present at the end of the electromagnetic spectrum.<\/h3>\n

2. 1057MHz1057MHz
\n(Frequency of Radiation Arising from Two Close Energy Levels in Hydrogen; Known as Lamb Shift).<\/h2>\n

Ans: Frequency of
\n1057MHz1057MHz
\nbelongs to radio waves. As, radio waves generally belong to the frequency range of 500kHz500kHz to
\n1000MHz1000MHz
\n. The range of wavelength for radio waves is, >0.1m>0.1m.<\/h3>\n

3. 2.7K2.7K
\nTemperatureAssociatedWiththeIsotropicRadiationFillingAllSpace\u2212ThoughttoBeaRelicofthe\u2018Big\u2212Bang\u2032OriginoftheUniverseTemperatureAssociatedWiththeIsotropicRadiationFillingAllSpace\u2212ThoughttoBeaRelicofthe\u2018Big\u2212Bang\u2032OriginoftheUniverse.<\/h2>\n

Ans: Given, temperature,
\nT=2.7KT=2.7K
\n.
\nAccording to Planck’s law,
\n\u03bb=0.29(cm)T(K)\u03bb=0.29(cm)T(K)
\nSubstituting the value of temperature, we get: \u03bb=0.292.7=0.11cm\u03bb=0.292.7=0.11cm.
\nNow,
\n\u03bb=0.11cm\u03bb=0.11cm
\nbelongs to the Microwave region of electromagnetic spectrum.<\/h3>\n

4. 5890A\u2218\u22125896A\u22185890A\u2218\u22125896A\u2218 doublelinesofsodiumdoublelinesofsodium<\/h2>\n

Ans: This wavelength belongs to yellow light of visible spectrum.<\/h3>\n

5. 14.4 keV14.4 keV
\n[energy of a particular transition in
\n57Fe57Fe
\nnucleus Associated With a Famous High Resolution Spectroscopic Method (Mossbauer Spectroscopy)].<\/h2>\n

Ans: Given: 14.4keV14.4keV
\nThe transition energy can be given as: E=h\u03bdE=h\u03bd.
\nWhere,
\nhh is Planck’s constant ,
\n\u03bd\u03bd is frequency of radiation.
\nSubstituting the values, we get:
\n\u03bd=Eh=14.4\u00d7103\u00d71.6\u00d710\u2212196.6\u00d710\u221234=3.4\u00d71018Hz\u03bd=Eh=14.4\u00d7103\u00d71.6\u00d710\u2212196.6\u00d710\u221234=3.4\u00d71018Hz.
\nNow, 3.4\u00d71018Hz3.4\u00d71018Hz belongs to range of X-ray frequencies. As, range of frequency for X-rays lies between
\n1016Hz\u22121020Hz1016Hz\u22121020Hz
\n.<\/h3>\n

15. Answer the Following Questions:
\n1. Long-Distance Radio Broadcasts Use Short-Wave Bands. Why?<\/h2>\n

Ans: Short wave bands are used for long distance broadcasting because, after getting reflected from ionosphere present in our environment, these waves can reach to longer distances.<\/h3>\n

2. It Is Necessary to Use Satellites for Long Distance Tv Transmission. Why?<\/h2>\n

Ans: Yes, for long-distance TV transmission, it is compulsory to use satellites, because TV signal contains very high frequency of range
\n54MHz\u2212890MHz54MHz\u2212890MHz
\n, and this much high frequency does not get reflected by ionosphere, and communication does not get stablished. Hence, we need to use satellites for TV transmission, over long distance.<\/h3>\n

3. Optical and Radio Telescopes Are Built on the Ground but X-Ray Astronomy Is Possible Only from Satellites Orbiting the Earth. Why?<\/h2>\n

Ans: Light and radio waves used in optical and radio telescopes respectively, penetrate through the atmosphere but X-rays used in X-ray astronomy, are absorbed by the atmosphere. Hence, optical and radio telescopes are built on earth’s surface and for X-ray astronomy, satellites orbiting the earth are necessary.<\/h3>\n

4. The Small Ozone Layer on Top of the Stratosphere is Crucial for Human Survival. Why?<\/h2>\n

Ans: The ozone layer, present on the top of the stratosphere layer is essential for human survival because it prevents the harmful radiations, (like, ultraviolet rays, \u03b3\u03b3 rays, cosmic rays) coming from the sun to earth’s atmosphere, which can cause many diseases like cancer and genetic damage in cells and plants. The ozone layer also traps the infrared radiations and maintains earth’s warmth.<\/h3>\n

5. If the Earth Did Not Have an Atmosphere, Would Its Average Surface Temperature Be Higher or Lower Than What it Is Now?<\/h2>\n

Ans: If the earth would not have any atmosphere, then there will not
\nbe any greenhouse effect. And if the greenhouse effect would not be present, then infrared radiations and CO2CO2would not retain in the atmosphere. The infrared radiations and CO2CO2 gets reflected from the atmosphere and keep the earth’s surface warm. But in the absence
\nof an atmosphere, this would not be possible and the average surface
\ntemperature of the earth would become lower.<\/h3>\n

6. Some Scientists Have Predicted That a Global Nuclear War on the Earth Would Be Followed by a Severe \u2018nuclear Winter\u2019 With a Devastating Effect on Life on Earth. What Might Be the Basis of This Prediction?<\/h2>\n

Ans: A global nuclear war on the earth would be followed by a severe \u2018nuclear winter\u2019, as the nuclear weapons would create clouds of smoke and dust, which would stop the solar light to reach the earth’s surface and would stop the greenhouse effect. This would cause extreme winter conditions.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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