Question

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

Answer 1

Correct Answer - C
`KE={{:(2E_(0)-E_(0)=E_(0).,,"for "0lexle1),(" "2E_(0).,,"for "x gt 1):}`
`thereforelambda_(1)=(h)/(sqrt(2mE_(0)))" and "lambda_(2)=(h)/sqrt(4mE_(0))implies (lambda_(1))/(lambda_(2))=sqrt(2)`