If the cathode-anode potential difference in an X-ray tube be $10^{5}$ V, then the maximum energy of X-ray photon can be

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$10^{5}$ J

$10^{5}$ MeV

$10^{- 1}$ MeV

$10^{5}$ KeV

answer is C.

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Detailed Solution

$\begin{array}{l} since λ_{\min} = \frac{12375}{V} \overset{o}{A} = \frac{12375}{10^{5}} \overset{o}{A} = 0 .123 \overset{o}{A} \\ E_{\max} = \frac{hc}{λ_{\min}}; \\ On putting the values . E_{\max} ≅ 10^{- 1} MeV . \end{array}$

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