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A particle of charge `q` and mas `m` si accelerated from set through a potential difference `V`. Its de Broglie wavelength is equal toA. `sqrt((h)/(2mq V))`B. `(hqV)/(sqrt(2m))`C. `sqrt((hqV)/(2m))`D. `(h)/(sqrt(2mqV))` |
Answer» Correct Answer - D When a charge `q` is accelerated from rest through a potential difference `V`, its gain in kinetic enegry `(1)/(2)mV^(2)` must equal its loss in potential enegry `qV`, since the energy is conserved. That is, `(1)/(2)mV^(2) = qV` Since momentum, `p = mV`, we can express this in the form `(1)/(2)(m^(2)v^(2))/(m) = qV rarr (p^(2))/(2m) = qV` or `p = sqrt(2mqV)` Substituting this expression for `p` into the de Broglie relation `(lambda = h//p)` gives `lambda = (h)/(sqrt(2mqV))` |
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