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The potential energy of a partical varie...

The potential energy of a partical varies as .
`U(x) = E_0 ` for ` 0 le x le 1`
`= 0` for `x gt 1 `
for `0 le x le 1` de- Broglie wavelength is `lambda_1` and for `xgt1` the de-Broglie wavelength is `lambda_2`. Total energy of the partical is `2E_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:
C
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