A charged particle is accelerated from rest through a certain potenetial difference. The de-Broglie wavelength is `lambda_(1)` when it is acceleated through `V_(1)` and is `lambda_(2)` when accelerated through `V_(2)`. Then ratio `lambda_(1)//lambda_(2)` is

Options

A. `V_(1)^(3//2) : V_(2)^(3//2)`
B. `V_(2)^(1//2):V_(1)^(1//2)`
C. `V_(1)^(1//2):V_(2)^(1//2)`
D. `V_(1)^(2):V_(2)^(2)`

Answer

B

Explanation

Wavelength , `lambda=(h)/(sqrt(mqv))`
`lambda_(1)(h)/(sqrt(m_1 qv_1)),lambda_(2)=(h)/(sqrt(m_2qv_2))`
`(lambda_1)/(lambda_2)=(sqrt(v_2))/(sqrt(v_1))`
`=V_(2)^((1)/(2)):V_(1)^((1)/(2))`.