Q122 / 125 · BITSAT 2009 · Physics
An X-ray tube is operated at $15 \mathrm{kV}$. Calculate the upper limit of the speed of the electrons striking the target.
- Correct answerA$7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}$
- B$7.62 \times 10^{7} \mathrm{~m} / \mathrm{s}$
- C$7.62 \times 10^{7} \mathrm{~cm} / \mathrm{s}$
- D$7.26 \times 10^{9} \mathrm{~m} / \mathrm{s}$
Solution
The maximum kinetic energy of an electron accelerated through a potential difference of $\mathrm{V}$ volt is $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$
$$
\begin{array}{l}
\therefore \text { maximum velocity } \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\
\mathrm{v}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 15000}{9.1 \times 10^{-31}}} \\
\mathrm{v}=7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}
\end{array}
$$
Hence, the correct answer is option: 'A'.