Ncert Solutions Physics Class 12th

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

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 -rays of equal energy. What is the wavelength associated with each -ray? (1BeV = 10⁹ eV)

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.24 The total energy of two X-rays = 10.2 BeV = 10.2 $\times 10^{9} e V$ = 10.2 $\times 10^{9} \times 1.6 \times 10^{- 19}$ J

Hence energy of each X-ray $E$ = $\frac{10.2 \times 10^{9} \times 1.6 \times 10^{- 19}}{2}$ = 8.16 $\times 10^{- 10}$ J

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Speed of light, c = 3 $\times 10^{8}$ m/s

From the relation of energy and wavelength, we get

$E$ = $\frac{h c}{λ}$ &nbs

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

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?

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.23 Wavelength produced by X-ray, $λ =$ 0.45 Å = 0.45 $\times 10^{- 10}$ m

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Speed of light, c = 3 $\times 10^{8}$ m/s

The maximum energy of a photon is given as:

$E$ = $\frac{h c}{λ}$ = $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{0.45 \times 10^{- 10}}$ = 4.417 $\times 10^{- 15}$ J = $\frac{4.417 \times 10^{- 15}}{1.6 \times 10^{19}}$ eV = 27.6 $\times 10^{3}$ eV = 27.6 keV

Theref

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

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 ( 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.

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.22 Potential, V = 100 V

Magnetic field experienced by electron, B = 2.83 $\times 10^{- 4}$ T

Radius of the circular orbit, r = 12.0 cm = 12 $\times 10^{- 2}$ 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.

$\frac{1}{2}$ m&nb

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

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 × 10¹¹C 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 emphasize 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.]

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.21 Speed of the electron, v = 5.20 $\times 10^{6} m / s$

Magnetic field experienced by the electron, B = 1.30 $\times 10^{- 4}$ T

Specific charge of electron, e/m = 1.76 $\times 10^{11}$ C/kg

Charge of an electron e = 1.60 $\times 10^{- 19}$ C

Mass of electron, m = 9.1 $\times 10^{- 31}$ kg

The force exerted on the electron is given as

F = $e |\overset{?}{v} + \overset{?}{B}|$

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

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 a collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.20 Potential difference across evacuated tube, V = 500 V

Specific charge of electron, e/m = 1.76 $\times 10^{11}$ C/kg

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ = $e V$

$v = \sqrt{\frac{2 e V}{m}}$ = $\sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 \times 500 \times 1.76 \times 10^{11}}$ = 13.27 $\times 10^{6}$ m/s

Therefore, the speed of ea

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

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)

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $\times$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $\times 10^{- 27}$ kg

So, m = 28.0152 $\times$ 1.66 $\times 10^{- 27}$ kg = 4.65 $\times 10^{- 26}$ kg

Planck’s constant, h = 6.626 $\times 10^{- 34}$ J

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

11.18 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

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Payal Gupta | Contributor-Level 10

a year ago

11.18 The momentum of a photon having energy ( $h ν)$ is given as $p$ = $\frac{h ν}{c}$ = $\frac{h}{λ}$

So, $λ = \frac{h}{p}$ …………….(1)

Where, $λ =$ Wavelength of the electromagnetic radiation

h = Planck’s constant

c = speed of light

De Broglie wavelength of the photon is given as $λ = \frac{h}{m v}$

But momentum, $= m v$ &nb

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

11.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10^–10 m?

(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.17 De Broglie wavelength of the neutron, $λ$ = 1.40 $\times 10^{- 10}$ m

Mass of neutron, m = 1.66 $\times 10^{- 27}$ kg

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Kinetic energy, $E_{k}$ = $\frac{1}{2} m v^{2}$ ………….(1)

De Broglie wavelength and velocity (v) are related as

$λ = \frac{h}{m v}$ …………….(2)

Combining equation (1) and (2

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter News

11.16 An electron and a photon each have a wavelength of 1.00 nm. Find

(a) Their momentum,

(b) The energy of the photon, and

(c) The kinetic energy of electron.

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.16 Wavelength of electron, $λ_{e}$ = Wavelength of proton, $λ_{p}$ = 1.0 nm = 1 $\times 10^{- 9}$ m

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

From De Broglie wavelength relation, $λ = \frac{h}{p},$ where p = momentum

$p$ = $\frac{h}{λ}$ = $\frac{6.626 \times 10^{- 34}}{1 \times 10^{- 9}}$ = 6.626 $\times 10^{- 25}$ kg.m/s. Since $λ_{e} = λ_{p}$ , their momentu

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Dual Nature of Radiation and Matter

11.15 What is the de Broglie wavelength of

(a) Bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,

(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and

(c) A dust particle of mass 1.0 × 10^–9 kg drifting with a speed of 2.2 m/s?

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

11.15 Mass of the bullet, m = 0.04 kg

Speed of the bullet, v = 1.0 km/s = 1000 m/s

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

De Broglie wavelength of the bullet is given by the relation:

$λ = \frac{h}{m \times v}$ = $\frac{6.626 \times 10^{- 34}}{0.04 \times 1000}$ = 1.65 $\times 10^{- 35}$ m

Mass of the ball, m = 0.06 kg

Speed of the ball, v = 1.0 m/s

Planck’s constant,

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Ncert Solutions Physics Class 12th

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