JEE Advanced 2023 Paper 2 · Physics · Atoms, Nuclei and Dual Nature of Radiation · hard

Zinc (Z=30) is used as the target of an X-ray tube. Its Kα photon energy may be taken as E=10.2 (Z-1)^2 eV. These photons fall on a thin metal foil…

Single correct option

Zinc (

Z=30

) is used as the target of an X-ray tube. Its

K_\alpha

photon energy may be taken as

E=10.2\,(Z-1)^2\ \mathrm{eV}

. These photons fall on a thin metal foil of work function

4.5\ \mathrm{eV}

, and the fastest photoelectrons enter a uniform magnetic field

B=1.0\ \mathrm{mT}

perpendicular to their velocity. Use

m_e=9.109\times10^{-31}\ \mathrm{kg}

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

. The radius of the circular path of these electrons is nearest to

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Answer: (B) $0.31\ \mathrm{m}$

Photon energy:

E=10.2\times29^2=8578.2\ \mathrm{eV}

. Maximum kinetic energy:

K=E-\phi=8578.2-4.5=8573.7\ \mathrm{eV}

(non-relativistic, since

K\ll m_ec^2

). Momentum

p=\sqrt{2m_eK}

with

K=8573.7\times1.602\times10^{-19}\ \mathrm{J}

gives

p\approx5.0\times10^{-23}\ \mathrm{kg\,m\,s^{-1}}

. In a magnetic field

r=\dfrac{p}{eB}=\dfrac{5.0\times10^{-23}}{1.602\times10^{-19}\times10^{-3}}\approx0.31\ \mathrm{m}

. The value

0.22\ \mathrm{m}

arises from using

p=\sqrt{m_eK}

(dropping the factor 2),

0.36\ \mathrm{m}

from using

13.6\,(Z-1)^2

for the

K_\alpha

energy (forgetting the factor

3/4

), and

0.62\ \mathrm{m}

is the diameter.

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