Question

An `alpha`-particle when accelerated through a potential of V volt has a wavelength `lambda` associated with it, but if a proton in order to have same wavelength `lambda` by what potential difference it must be accelerated?
A. 8 V
B. 6 V
C. 4 V
D. 12 V

Answer 1

Correct Answer - A
`lambda_(p)=lambda_(alpha)`
`(h)/(sqrt(2m_(p)Q_(p)V_(p)))=(h)/(sqrt(2m_(alpha)Q_(alpha)V_(alpha)))`
`therefore" "m_(p)Q_(p)V_(p)=m_(alpha)Q_(alpha)V_(alpha)`
`therefore" "V_(p)=((m_(alpha))/(m_(p)))((Q_(alpha))/(Q_(p)))V_(alpha`
`V_(p)=(4)(2)V_(alpha)implies V_(p)=8V`