A particle is dropped 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

From equation `v^(2)-v_(0)^(2)=2ad` of uniformly accelerated motion,here for freely falling particle ,taking `v_(0),a=-g` and d=-H ,
`v^(2)-(0)^(2)=2(-g)(-H)`
`therefore v^(2)=2gh`…..(1)
`therefore v=sqrt(2gh)` (velocity of particle when it is aout to touch the ground)
Now,according to formula,when above particle is about to touch the ground ,de-Broglie wavelength of a wavw ,representing that particle is,
`lmbda=(h)/(mv)` (where h=plank.s constant)
m=mass of a particle
v=velocity of particle )
`=(h)/(msqrt(2gh))`(from equation(1))
`lambda prop(1)/(sqrt(H))` (`therefore` h,m and g are constants)
`therefore lambda prop H^((-1)/(2))`
`implies` Option (D) is correct.