An electron of mass `m_e` and a proton of mass `m_p` are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

Options

A. 1
B. `(m_(p))/(m_(e))`
C. `(m_(e))/(m_(p))`
D. `sqrt((m_(p))/(m_(e))`

Answer

D

Explanation

If q is the charge on the particle and V the potential difference through which it is accelerated then, `qV=(1)/(2)mv^(2)`
or `mv=sqrt(2mqV)`
de Broglie's wavelength,
`lamda=(h)/(mv)=(h)/(sqrt(2mqV)`
`(lamda_e)/(lamda_p)=sqrt((m_p)/(m_e))`