A particle is droped from a height H. The de-broglie wavelength of the particle as a function of height is proportional to

Options

A. H
B. `H^(1//2)`
C. `H^0`
D. `H^(-1//2)`

Answer

D

Explanation

When the body is dropped from a height H, it acquires a velocity `v=sqrt(2gh) [ therefore v^2 = u^2 + 2gh]`
`lamda = (h)/(mv) = (h)/(m(sqrt(2gH))) = (h)/(msqrt(2g)sqrtH)`
In this case `((h)/(msqrt(2g)))` is a constant say K
`therefore lamda = K/sqrtH " or " lamda prop 1/sqrtH " or " lamda prop H^(-1//2)`
Note that H is the height and h is planck's constant .