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When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)`C
Q. The speed of the particle that can take discrete values is proportional to

A

`n^(-32)`

B

`n^(-1)`

C

`n^(1//2)`

D

`n`

Text Solution

Verified by Experts

The correct Answer is:
D

`P = (n h)/(2a) = m v`
`rArr v prop n`
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The relation between kinetic energy K and linear momentum p of a particle is represented by

RESONANCE-ATOMIC PHYSICS-Exercise -3 part -I JEE (Advanced)
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