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).`

Options

A. `sqrt(2)`
B. `(1)/(sqrt(2))`
C. 2
D. `(1)/(2)`

Answer

A

Explanation

For `0 le x le 1, KE = 2E_(0) - E_(0) =E_(0)`
for `x gt 1`
`(lamda_(1))/(lamda_(2))=(h//P_(1))/(h//P_(2))`
`=(P_(2))/(P_(1))=sqrt((KE_(2))/(KE_(1)))`
`= sqrt((2E_(0))/(E_(0)))=sqrt(2)`