Q:

What is the deBroglie Wavelength in a region with potential energy?

This question was on a physics GRE:
A free particle with initial kinetic energy $T$ and deBroglie wavelength $\lambda$ enters a region in which it has potential energy $V$. what is the particle's new deBroglie wavelength?
My solution: The deBroglie wavelength is $\lambda = h/p$. Since in general $\frac{p^2}{2m}=T$ (where $T$ is kinetic energy only), the wavelength is only dependent on kinetic and not potential energy, so $\lambda '=\lambda$. But this is not an answer choice. My first question: why is this wrong?
Solutions on the internet use the fact that "the energy of the particle when it enters the new region is $E_f = T - V$", where $T$ is the kinetic energy the particle started with. If I assume this I can solve the problem, despite not knowing why my solution was wrong. My second question: If the Hamiltonian is $H=T+V$ in the Schrödinger equation, then why is it not $E_f = T + V$?
If the energy should in fact be $T+V$, then what is a valid solution to this? The answer is supposed to be $\lambda '=\lambda (1-V/T)^{-1/2}$ according to the official answer key by ETS. 

A:

I think you're reading too much into this. You are intended to treat the particle as an infinite plane wave moving with some initial energy $T$ and some final energy $T-V$. That is, the particle has to do some work $V$ to climb into the region of increased potential energy, and that work comes from its kinetic energy.
The rest is just algebra. Start with:
$$ \lambda = \frac{h}{\sqrt{2mT}} $$
Then:
$$\begin{align}
 \lambda' &= \frac{h}{\sqrt{2m(T-V)}} \\
          &= \frac{h}{\sqrt{2mT}} \frac{1}{\sqrt{1-V/T)}}  \\
          &= \frac{\lambda}{\sqrt{1-V/T)}} 
\end{align}$$

