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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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