Q101 / 125 · BITSAT 2015 · Physics
When $0.50 Ã$ X-rays strike a material, the photoelectrons from the $\mathrm{k}$ shell are observed to move in a circle of radius $23 \mathrm{~mm}$ in a magnetic field of $2 \times 10^{-2}$ tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?
- A$3.5 \mathrm{keV}$
- Correct answerB$6.2 \mathrm{keV}$
- C$2.9 \mathrm{keV}$
- D$5.5 \mathrm{keV}$
Solution
As we know,
$$
\mathrm{F}=\mathrm{qvB}=\mathrm{m} \frac{\mathrm{v}^{2}}{\mathrm{R}} \Rightarrow \mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{BR}
$$
The kinetic energy of the photoelectron
$$
\begin{array}{l}
=\frac{1}{2} m v^{2}=\frac{1}{2} \frac{e^{2} B^{2} R^{2}}{m} \\
=\frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)^{2}\left(2 \times 10^{-2}\right)^{2}\left(23 \times 10^{-3}\right)^{2}}{\left(9.1 \times 10^{-31}\right)} \\
=2.97 \times 10^{-15} \mathrm{~J} \\
=\frac{2.97 \times 10^{-15}}{1.6 \times 10^{-19}}=18.36 \mathrm{keV}
\end{array}
$$
Fnergy of the incident $\mathrm{n}=\frac{\mathrm{hc}}{\lambda}$
$$
=\frac{12.4}{0.50}=24.8 \mathrm{keV}
$$
Therefore, Binding energy $=24.8-18.6=$ $6.2 \mathrm{keV}$
Hence, the correct answer is option: 'B'.