An `alpha`- particle and a proton are accelerated from rest by a potential difference of `100 V` After this their de Broglie wavelength are `lambda_(a) and lambda_(p)` respectively The ratio `(lambda_(p))/ lambda_(p)` , to the nearest integer is

Options

A. 3
B. 4
C. 2
D. 45

Answer

B

Explanation

The de-Broglie wavelength of an electron is given by
`lamda=(h)/(mv)=(h)/(sqrt(2meV))=(12.27)/(sqrt(V))` Å (h=Planck's constant)
where, m=mass of electron, e=electronic charge and
V=potential difference with which electron is accelerated
`lamda=(12.27)/(sqrt(200))"Å"=(12.27)/(10)"Å"=1.227` Å