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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_0,0lexle1),(0," "x gt1):}` `lambda_1` and `lambda_2`​ are the de-Broglie wavelengths of the particle, when `0 le x le 1` and x > 1 respectively. If the total energy of particle is `2E_0`. find `(lambda_1//lambda_2)^2`

A

1

B

2

C

3

D

`1//2`

Text Solution

Verified by Experts

The correct Answer is:
B
K.E.= `2E_0-E_0=E_0` (for `0 le x le 1`)
`lambda_1=h/sqrt(2mE_0)`
KE=`2E_0` (for x > 1),
`lambda_2=h/sqrt(4mE_0), lambda_1/lambda_2=sqrt2`
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