1.

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

Answer»

H
`H^((1)/(2))`
`H^(0)`
`H^((-1)/(2))`

SOLUTION :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.


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