Ans : (b)
Total energy = Kinetic energy + Potential energy
For the region $ 0 \leq x \leq 1,$
Kinetic energy = Total energy – Potential energy
$= 2E_{\circ} – E_{\circ} = E_{\circ}$
So, $ \lambda_1 = \large\frac{h}{ \sqrt {2m(E_{\circ})}}$…………..(i)
For the region $x>1$,
Kinetic energy = Total energy – Potential energy
$= 2E_{\circ} – 0 =2 E_{\circ}$
So, $ \lambda_2 = \large\frac{h}{ \sqrt{2m(2E_{\circ})}}$…………….(ii)
From (i) & (ii),
$ \large\frac{ \lambda_1}{ \lambda_2} = \sqrt 2$