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).`

Answer

`sqrt(2)`

Explanation

For `0 lt x le1, " "KE=2E_(0)-E_(0)=E_(0)`
for `x gt 1," " KE=2E_(0)" "(lambda_(1))/(lambda_(2))=(h//P_(1))/(h//P_(2))=(P_(2))/(P_(1))=sqrt((KE_(2))/(KE_(1)))=sqrt(2)`