When 0.50 ; A ; X-rays strike a material, the photoelectrons from… - Vidyadip

MCQ

When $0.50\; \mathring A\; X$-rays strike a material, the photoelectrons from the $k$ shell are observed to move in a circle of radius $23\,mm$ in a magnetic field of $2 \times 10^{-2}$ tesla acting perpendicularly to the direction of emission of photoelectrons.$..........\, keV$ is the binding energy of $k$-shell electrons.

✓
$3.5$
B
$6.2$
C
$2.9$
D
$5.5$

✓

Answer

Correct option: A.

$3.5$

a
(b)

The velocity of the photoelectrons is found by the relation :

$e v B=m \frac{v^2}{R} \text { or } v=\frac{e}{m} B R$

The kinetic energy of the hotoelectrons is

$K =\frac{1}{2} m v^2=\frac{1}{2} \frac{e^2 B^2 R^2}{m}$

$=\frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)^2\left(2 \times 101^{-2}\right)^2\left(23 \times 10^{-3}\right)^2}{\left(9.1 \times 10^{-31}\right)}$

$=2.97 \times 10^{-15}\,J$

$=\left(2.97 \times 10^{-15}\right) \frac{1}{1.6 \times 10^{-19}}=18.36\,KeV$

The energy of the incident photon is $E_v=\frac{h c}{\lambda}=\frac{12.4}{0.50}=24.8\,KeV$

The binding energy is the difference between these two values:

$B E=E_v-K=24.8 - 18.6- 6.2\,keV$

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