A proton when accelerated through a potential difference of V, has a De-Broglie wavelength `lambda` associated with it.if an `alpha`-particle is to have the same De-Broglie wavelength `lambda,` it must be accelerated through a potential difference of

Options

A. `V/8`
B. `V/4`
C. `4 V`
D. `8 V`

Answer

a

Explanation

`lambda_p=lambda_prop
(mqV)_p=(mqV)_prop`
Potential difference,`vprop=V/8(therefore m_prop=4m_p,2q_p)`