Atoms, Nuclei and Dual Nature of…

JEE Advanced 2023 Paper 2 Physics: Paragraph 1

X-rays of wavelength

5.00\ \mathrm{pm}

are scattered by free electrons initially at rest (Compton scattering). The scattered photons are observed at an angle of

60^\circ

to the incident direction. Use

h=6.626\times10^{-34}\ \mathrm{J\,s}

c=3.00\times10^{8}\ \mathrm{m\,s^{-1}}

m_e=9.11\times10^{-31}\ \mathrm{kg}

and

e=1.602\times10^{-19}\ \mathrm{C}

. Relativistic corrections to the electron's kinetic energy are to be handled through conservation of energy and momentum of the photon-electron system.

Question 1

Numerical answer

Find the kinetic energy of the recoiling electron in

\mathrm{keV}

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 48.43

Compton shift:

\Delta\lambda=\dfrac{h}{m_ec}(1-\cos\theta)

with

\dfrac{h}{m_ec}=\dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times3\times10^{8}}=2.4244\ \mathrm{pm}

. For

\theta=60^\circ

\Delta\lambda=1.2122\ \mathrm{pm}

and

\lambda'=6.2122\ \mathrm{pm}

. By energy conservation the electron's kinetic energy is the photon's loss:

K=hc\left(\dfrac1\lambda-\dfrac1{\lambda'}\right)

. With

hc=1.9878\times10^{-25}\ \mathrm{J\,m}

\dfrac{hc}{\lambda}=3.9756\times10^{-14}\ \mathrm{J}=248.2\ \mathrm{keV}

and

\dfrac{hc}{\lambda'}=199.7\ \mathrm{keV}

. Hence

K\approx48.43\ \mathrm{keV}

. (Checking with momentum: the electron momentum from the components gives

K=\sqrt{p^2c^2+m^2c^4}-mc^2

with the same value.)

Question 2

Numerical answer

Find the angle in degrees that the recoil electron's direction makes with the direction of the incident photons. Give the answer correct to two decimal places.

Show answer and solution

Answer: 49.39

Momentum conservation. Along the incident direction:

p_e\cos\varphi=\dfrac h\lambda-\dfrac h{\lambda'}\cos60^\circ

. Perpendicular:

p_e\sin\varphi=\dfrac h{\lambda'}\sin60^\circ

(electron goes to the opposite side of the scattered photon). Dividing,

\tan\varphi=\dfrac{\sin60^\circ/\lambda'}{1/\lambda-\cos60^\circ/\lambda'}

. With

\lambda=5.000\ \mathrm{pm}

and

\lambda'=6.2122\ \mathrm{pm}

: numerator

=0.8660/6.2122=0.13940

, denominator

=0.20000-0.5/6.2122=0.11951

. Therefore

\tan\varphi=1.1664

and

\varphi\approx49.39^\circ

. This is consistent with the energy found earlier: the magnitude

p_e

yields the same kinetic energy.

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