## University Physics with Modern Physics (14th Edition)

$\sqrt{\frac{1}{2}}$.
Outside the barrier, for $x \gt L$, $U=0$ and the momentum $p=\sqrt{2m(E-U_o)}= \sqrt{2m(2U_o)}$. Inside the barrier, for $0 \lt x \lt L$, $U=0$ and the momentum $p=\sqrt{2m(E- U_o)}= \sqrt{2m(U_o)}$. The de Broglie wavelength is inversely proportional to the momentum, $\lambda=\frac{h}{p}$. The ratio of wavelengths is now calculated. $$\frac{\lambda_{out}}{\lambda_{in}}=\frac{\sqrt{2m(U_o)}}{ \sqrt{2m(2U_o)}}$$ $$\frac{\lambda_{out}}{\lambda_{in}}=\sqrt{\frac{1}{2}}$$