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BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Physics
Chapter: Electromagnetic Waves

BITSAT 2009 · Physics · Electromagnetic Waves

An X-ray tube is operated at 15 kV . Calculate the upper limit of...

An X-ray tube is operated at

15 \mathrm{kV}

. Calculate the upper limit of the speed of the electrons striking the target.

Options

A $7.26 \times 10^{7} \mathrm{ m} / \mathrm{s}$
B $7.62 \times 10^{7} \mathrm{ m} / \mathrm{s}$
C $7.62 \times 10^{7} \mathrm{ cm} / \mathrm{s}$
D $7.26 \times 10^{9} \mathrm{ m} / \mathrm{s}$

Correct answer

A. $7.26 \times 10^{7} \mathrm{ m} / \mathrm{s}$

Step-by-step 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 \times10^{7} \mathrm{ m}/ \mathrm{s} \end{array}

$$

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