A proton accelerated through a potential `V` has de-Broglie wavelength `lamda`. Then, the de-Broglie wavelengthof an alpha-particle, when accelerated through the same potential `V` is

Options

A. `(lamda)/2`
B. `(lamda)/(sqrt(2))`
C. `(lamda)/(2(sqrt(2))`
D. `(lamda)/8`

Answer

C

Explanation

The de-Broglie wavelength of a proton when accelerated through potential `V` is `lamda=h/(sqrt(2mqV))`
Now, the de-Broglie wavelength of `alpha`- particle, when accelerated through `V` is
`lamda'=h/(sqrt(2m'q'v'))`
where `m'=4m` and `q'=2q`
`lamda'=h/(sqrt(2xx4mxx2qV))=(lamda)/(2sqrt(2))`