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 xgt the de-Broglie wavelength is `lambda_2`. Total energy of the partical is 2E_0. find `(lambda_1)/(lambda_2).`

Explanation

`For 0lexle 1, PE = E_0`
. :.Kinetic energy K_1 = Total energy - PE
`:.lambda_1 = (h)/(sqrt2me_0) …(i)`
. For xgt, PE = 0
. :. Kinetic energy `K_2` = Total energy `= 2E_0`
. `:.lambda_2 = (h)/(sqrt4mE_0) …(ii)`
. From Eqs (i) and(ii), we have `lambda_1)/(lambda_2) = sqrt2.`