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

The de-Broglie wavelength associated with a charged particle is given by
`lambda = (h)/(sqrt(mqV))`
Where, h = planck's constant
m = mass of charged particle
q = charge on the particle
V = potential at which the charged particle is accelerated
`:. lambda_(1) = (h)/(sqrt(m_(1) q v_(1))) lambda_(2) = (h)/(sqrt(m_(2)qv_(2)))`
Here, `m_(1) = m_(2)`
`implies `(lambda_(1))/(lambda_(2)) = sqrt((m_(2) q v_(2))/(m_(1) q v_(1))) = sqrt((V_(2))/(V_(1)))`
`= V_(2)^((1)/(2)) . V_(1)^((1)/(2))`