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The potential energy of a particle of ma...

The potential energy of a particle of mass `m` is given by
`V(x) = E_0` when `x= lex le 1 `and `xgt 1` repectively.
`lambda_(1) and lambda_(2) ` are the de - Broglie wavelength of the particle, ,if the total energy of particle is `2 E_(0)` find `lambda_(1) // lambda_(2)`

A

2

B

1

C

`sqrt(2)`

D

`(1)/(sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
3
`K.E.=2E_(0)-E_(0) ("for"0lexle1)rArrlambda_(1) = (h)/(sqrt(4mE_(0))rArr (lambda_(1))/(lambda_(2))=sqrt(2).`
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