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If the cathode - anode potential difference in an `X` - ray tube be `10^(5) V` , then the maximum energy of `X` - ray photon can beA. `10^(5) J`B. `10^(5) MeV`C. `10^(-1) MeV`D. `10^(5) KeV` |
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Answer» Correct Answer - C Since `lambda_(min) = (12375)/(V) Å = (12375)/(10^(5)) Å = 0.123 Å` `E_(max) = ( hc) /(lambda_(min))`, On putting the values . `E_(max) ~= 10^(-1) MeV`. |
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