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

Velocity v, of freely falling body after falling from a height H will be: `H= v = sqrt(2gH)`. We know that de-Broglie wavelength, `lamda= (h)/(p)`
`lamda= (h)/(mv)= (h)/(m sqrt(2gH))` h, m and g are constant `(h)/(m sqrt(2g))` is constant `rArr lamda prop (1)/(sqrtH) rArr lamda prop H^(-1//2)`