The voltage applied to an X-ray tube is $18 k V$ . The maximum mass of photon emitted by the X-ray tube will be:

2 \times 10^{- 13} k g

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3.2 \times 10^{- 32} k g

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6.4 \times 10^{- 32} k g

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9.1 \times 10^{- 31} k g

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Solution

The correct option is C $6.4 \times 10^{- 32} k g$
$e V = \frac{1}{2} m v^{2}$
or
$1.6 \times 10^{- 19} \times 18 \times 10^{3} = \frac{1}{2} m (3 \times 10^{8})^{2}$
or
$m = 6.4 \times 10^{- 32} k g$

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10^{5}

The minimum wavelength of X-rays emitted from an X-ray tube operating at a voltage of $10^{4} V$ is roughly equal to

20 k V

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19.9 K e V

A^{0}

= 6.63 \times 10^{- 34}

c = 3 \times 10^{8}

1 A^{o}

h = 6.6 \times 10^{- 34} J s

e = 1.6 \times 10^{- 19} C

c = 3 \times 10^{8} m / s

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