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 le x le 1, PE=E_(0)`
`:.` Kinetic energy `K_(1)=` Total energy `-PE`
`=2E_(0)-E_(0)=E_(0) :. Lambda_(1)=(h)/sqrt(2m(2E_(0)))` …(i)
For `x gt 1, PE =0`
`:.` Kinetic energy `K_(2)=` Total energy `=2E_(0)`
`:. lambda_(2) =(h)/sqrt(2m(2E_(0)))` ...(ii)
From equation (i) and (ii), we have `lambda_(1)/lambda_(2)=sqrt(2)`