The potential energy of a particle of mass m is given by `U(x){{:(E_(0),,,0lexlt1),(0,,,xgt1):}`
`lambda_(1)` and `lambda_(2)` are the de-Broglie wavelength of the particle, when `0le x le1` and `x gt 1` respectively. If the total energy of particle is `2E_(0)`, then the ratio `(lambda_(1))/(lambda_(2))` will be

Options

A. 1
B. 2
C. 1
D. `1//2`

Answer

B

Explanation

`K.E=2E_(0)-E_(0)=E_(0)` (for `0 let x le 1`)
`lamda_(1)=h/(sqrt(2mE_(0)))`
`lamda_(2)=h/(sqrt(mE_(0))),(lamda_(1))/(lamda_(2))=sqrt(2)`