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An X-ray tube is operated at $15 \mathrm{kV}$. Calculate the upper limit of the speed of the electrons striking the target.
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Verified Answer
The correct answer is:
$7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}$
The maximum kinetic energy of an electron accelerated through a potential difference of $\mathrm{V}$ volt is $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$
$$
\begin{array}{l}
\therefore \text { maximum velocity } \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\
\mathrm{v}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 15000}{9.1 \times 10^{-31}}} \\
\mathrm{v}=7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}
\end{array}
$$
$$
\begin{array}{l}
\therefore \text { maximum velocity } \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\
\mathrm{v}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 15000}{9.1 \times 10^{-31}}} \\
\mathrm{v}=7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}
\end{array}
$$
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