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# NCERT Solutions (Part - 2)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

## Class 12 : NCERT Solutions (Part - 2)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

The document NCERT Solutions (Part - 2)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev is a part of the Class 12 Course Physics Class 12.
All you need of Class 12 at this link: Class 12

Ques 11.19:
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Ans:  Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 × 14.0076 = 28.0152 u
But 1 u = 1.66 × 10−27 kg
m = 28.0152 ×1.66 × 10−27 kg
Planck’s constant, h = 6.63 × 10−34 Js
Boltzmann constant, k = 1.38 × 10−23 J K−1
We have the expression that relates mean kinetic energy of the nitrogen molecule with the root mean square speed as: Hence, the de Broglie wavelength of the nitrogen molecule is given as: Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.

Ques 11.20:
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg−1.

(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Ans:  (a)Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as: Therefore, the speed of each emitted electron is (b) Potential of the anode, V = 10 MV = 10 × 106 V
The speed of each electron is given as: This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c.

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:

E = mc2
Where,
m = Relativistic mass m0 = Mass of the particle at rest
Kinetic energy is given as:
K = mc2m0c2

Ques 11.21:
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1.

(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Ans:  (a)Speed of an electron, v = 5.20 × 106 m/s

Magnetic field experienced by the electron, B = 1.30 × 10−4 T

Specific charge of an electron, e/m = 1.76 × 1011 C kg−1

Where,
e = Charge on the electron = 1.6 × 10−19 C
m = Mass of the electron = 9.1 × 10−31 kg−1
The force exerted on the electron is given as: θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam. The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam.

Hence, equation (1) reduces to: Therefore, the radius of the circular path is 22.7 cm.

(b) Energy of the electron beam, E = 20 MeV

The energy of the electron is given as: This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c

When very high speeds are concerned, the relativistic domain comes into consideration.

In the relativistic domain, mass is given as: Where,

m= Mass of the particle at rest

Hence, the radius of the circular path is given as: Ques 11.22:
An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10−2 mm of Hg). A magnetic field of 2.83 × 10−4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method. Determine e/m from the data.
Ans:  Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B = 2.83 × 10−4 T
Radius of the circular orbit r = 12.0 cm = 12.0 × 10−2 m
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e., It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam. Hence, we can write:

Centripetal force = Magnetic force Putting the value of v in equation (1), we get: Therefore, the specific charge ratio (e/m) is Ques 11.23:
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Ans:  (a) Wavelength produced by an X-ray tube, Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
The maximum energy of a photon is given as: Therefore, the maximum energy of an X-ray photon is 27.6 keV.

(b) Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.

Ques 11.24:
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)
Ans: Total energy of two γ-rays:
E = 10. 2 BeV
= 10.2 × 109 eV
= 10.2 × 109 × 1.6 × 10−10 J
Hence, the energy of each γ-ray: Planck’s constant, Speed of light, Energy is related to wavelength as: Therefore, the wavelength associated with each γ-ray is Ques 11.25:
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.

(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.

(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10−10 W m−2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz.

Ans: (a) Power of the medium wave transmitter, P = 10 kW = 104 W = 104 J/s

Hence, energy emitted by the transmitter per second, E = 104

Wavelength of the radio wave, λ = 500 m

The energy of the wave is given as: Where,

h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s Let n be the number of photons emitted by the transmitter.

nE1 = E The energy (E1) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large.

The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.

(b) Intensity of light perceived by the human eye, I = 10−10 W m−2
Area of a pupil, A = 0.4 cm2 = 0.4 × 10−4 m2
Frequency of white light, ν= 6 × 1014 Hz
The energy emitted by a photon is given as:

E = hν
Where,
h = Planck’s constant = 6.6 × 10−34 Js
E = 6.6 × 10−34 × 6 × 1014
= 3.96 × 10−19 J
Let n be the total number of photons falling per second, per unit area of the pupil.
The total energy per unit for n falling photons is given as:

E = n × 3.96 × 10−19 J s−1 m−2

The energy per unit area per second is the intensity of light.
E = I
n × 3.96 × 10−19 = 10−10 = 2.52 × 108 m2 s−1
The total number of photons entering the pupil per second is given as:
nA = n × A
= 2.52 × 108 × 0.4 × 10−4
= 1.008 × 104 s−1
This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

Ques 11.26:
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105 W m−2) red light of wavelength 6328 Å produced by a He-Ne laser?

Ans: Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m
Stopping potential of the metal, V0 = 1.3 V
Planck’s constant, h = 6.6 × 10−34 J
Charge on an electron, e = 1.6 × 10−19 C
Work function of the metal = Frequency of light = ν
We have the photo-energy relation from the photoelectric effect as:

= eV0 Let ν0 be the threshold frequency of the metal.  Wavelength of red light,   = 6328 × 10−10 m
∴ Frequency of red light,   Since ν0> νr, the photocell will not respond to the red light produced by the laser.

Ques 11.27:

Monochromatic radiation of wavelength 640.2 nm (1nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Ans: Wavelength of the monochromatic radiation, λ = 640.2 nm
= 640.2 × 10−9 m

Stopping potential of the neon lamp, V0 = 0.54 V
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.6 × 10−34 Js
Let φo be the work function and ν be the frequency of emitted light.

We have the photo-energy relation from the photoelectric effect as:

eV0 = − φo Wavelength of the radiation emitted from an iron source, λ' = 427.2 nm

= 427.2 × 10−9 m

Let V'0 be the new stopping potential. Hence, photo-energy is given as: Hence, the new stopping potential is 1.50 eV.

Ques 11.28:
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

λ1 = 3650 Å, λ2= 4047 Å, λ3= 4358 Å, λ4= 5461 Å, λ5= 6907 Å,

The stopping voltages, respectively, were measured to be:

V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V

Determine the value of Planck’s constant h, the threshold frequency and work function for the material.

[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10−19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Ans: Einstein’s photoelectric equation is given as:

eV0 = − φo Where,

V0 = Stopping potential
h = Planck’s constant
e = Charge on an electron
φ= Work function of a material

It can be concluded from equation (1) that potential V0 is directly proportional to frequency ν.

Frequency is also given by the relation: This relation can be used to obtain the frequencies of the various lines of the given wavelengths. The given quantities can be listed in tabular form as:

 Frequency x 1014 Hz 8.219 7.412 6.884 5.493 4.343 Stopping potential V0 1.28 0.95 0.74 0.16 0

The following figure shows a graph between and V0. It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 × 1014 Hz, which is the threshold frequency (ν0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the λ5 line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line = From equation (1), the slope can be written as: The work function of the metal is given as:

φ= hν0

= 6.573 × 10−34 × 5 × 1014

= 3.286 × 10−19 J = 2.054 eV

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