We assume that the energies and momenta involved are nonrelativistic. The de Broglie wavelength is defined as
$\begin{align*}\lambda := \frac{h}{p} = \frac{h}{\sqrt{2 m K}}\end{align*}$
where $K$ is the kinetic energy of the particle. Initially, $K = E$ , but after the particle enters the region in which it has a potential energy $V$ , the kinetic energy becomes $K = E - V$ . Therefore, the initial de Broglie wavelength must be multiplied by a factor of
$\begin{align*}\sqrt{\frac{E}{E-V}} = \sqrt{\frac{1}{1 - V/E}}\end{align*}$
in order to obtain the new de Broglie wavelength. Therefore, the correct answer is (E).

Solution to 2001 Problem 46