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. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

Let charge q be accelerated through a potential difference V, then we can write the following :
`qV=1/2mv^2`
`impliesm^2v^2=2mqV`
`implies mv = sqrt(2mqV)`
The de Broglie wavelength can be written as follows :
`implieslamda=h/(mv)=h/(sqrt(2mqV))`
Ratio of de Broglie wavelength for the two particles can be written as follows :
`implieslamda_p/lamda_alpha=sqrt((m_alphaq_alpha)/(m_pq_p))`
`implieslamda_p/lamda_alpha=sqrt((4xx2)/(1xx1))=2sqrt2=2.828`