An electron and a proton are accelerated through the same potential difference. The ratio of their de broglic wavelengths will be

Options

A. `(m_p/m_e)^2`
B. `m_e/m_p`
C. `m_p/m_e`
D. 1

Explanation

`lamda= (h)/(mv) = (h)/(sqrt(2mqv))`
`therefore lamda_e/lamda_P = sqrt((m_Pq_PV)/(m_e xx e xx V)) = sqrt(m_P/m_e)`
`therefore` V is the same and `q_P` = e (in magnitude)
`therefore (lamda_e/lamda_P) = (m_P/m_e)^(1//2)`