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If the voltage applied to an X-ray tube is increased to 1.5 times the minimum wavelength $λ_{m i n}$ of an X-ray continuous spectrum shifts by $∆ λ = 26 p m$ . The initial voltage applied to the tube is

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≈ 75 kV

≈ 10 kV

≈ 50 kV

≈ 16 kV

answer is B.

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Detailed Solution

$λ_{m i n} = \frac{h c}{e V_{1}} = λ_{1} a n d λ_{2} = \frac{h c}{e} (\frac{1}{V_{2}})$

$∆ λ = \frac{h c}{e} (\frac{1}{V_{1}} - \frac{1}{V_{2}}) = 26 \times 10^{- 2} A$

Given V₂ = 1.5 V₁
on solving we get V₁= 16000 volt = 16 kV

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