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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
`K.E. = 2 E_(0) - E_(0) = E_(0) (for 0le x le 1)`
`rArr lambda_(1) = (h)/(sqrt( 2 m E_(0))`
`K.E. = 2 E_(0) ( for x gt 1) rArr lambda_(2) = (h)/(sqrt( 4 m E_(0))) rArr (lambda_(1))/(lambda_(2)) = sqrt(2)`
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