A potential difference of 60 KV is applied across the ends of an X-Ray tube. The energy of the most energetic X-Ray photon is

9.6 \times 10^{- 19} J

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9.6 \times 10^{- 15} J

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9.6 \times 10^{- 16} J

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9.6 \times 10^{- 17} J

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Solution

The correct option is B $9.6 \times 10^{- 15} J$
$E n e r g y m a x = p o t e n t i a l d i f f e r e n c e \times e$
= $60 \times 10^{3} \times 1.6 \times 10^{- 19}$
= $9.6 \times 10^{- 15} J$ .
So, the answer is option (B).

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Similar questions

Find the minimum wavelength of X-ray produced if 10 kV potential difference is applied across the anode and cathode of the tube.

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

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

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