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The potential energy of a particle of ma...

The potential energy of a particle of mass m is given by `U(x){{:(E_(0),,,0lexlt1),(0,,,xgt1):}`
`lambda_(1)` and `lambda_(2)` are the de-Broglie wavelength of the particle, when `0le x le1` and `x gt 1` respectively. If the total energy of particle is `2E_(0)`, then the ratio `(lambda_(1))/(lambda_(2))` will be

A

2

B

1

C

`sqrt(2)`

D

`1/(sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
`KE =2E_(0)-E_(0)=E_(0)" for " (0 le x le 1) rArr lambda_(1)=h/(sqrt(2mE_(0)))`
`KE =2E_(0)(x gt 1) rArr lambda_(2) =h/(sqrt(4mE_(0)))`
`(lambda_(1))/(lambda_(2))=sqrt(2)`
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