An X-Ray tube emits a wavelength of $1 A^{o}$ . The potential difference to be applied across the ends of the tube is
(given $h = 6.6 \times 10^{- 34} J s$ , $e = 1.6 \times 10^{- 19} C$ and $c = 3 \times 10^{8} m / s$ )

124 V

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1240 V

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12400 V

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10 KV

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Solution

The correct option is C 12400 V
$λ = 1 A^{\circ} = 0.1 n m$
$e n e r g y = \frac{h c}{λ}$

$= \frac{1240 e V - n m}{0.1 n m} (h c = 1240 e V - n m)$

$= 12400 e V$
$e n e r g y = p . d \times e$
$12400 e V = p . d \times e$
$p . d = 12400 V .$
So, the answer is option (C).

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