Answer
The rank of the three regions according to the de Broglie wavelength of the electron is $\lambda_{1}\gt \lambda_{3} \gt \lambda_{2}$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$, ......................$(1)$
where $h$ is the planck's constant and $p$ is the momentum of the moving electron.
Again, the relation between $p$ and $E$ is
$E=\frac{p^2}{2m}$ , where $m$ is the mass of the electron.
Thus, $p=\sqrt {2mE}$ ......................$(2)$
Substituting eq. $2$ in eq. $1$, we get
$\lambda=\frac{h}{\sqrt {2mE}}$ ......................$(3)$
If $V$ be the magnitude of the electric potential of any region, then the corresponding kinetic energy $K$ of the electron can be written as $K=-eV$, where $e$ is the charge of electron.
Now, in region 1,
total energy of electron= $K$ in region 1
or, $E_{1}=100e$ $J$
In region 2,
total energy of electron= $K$ in region 1+ $K$ in region 2
or, $E_{2}=(100e +200e)$$J$ $=300e$ $J$
In region 3,
total energy of electron= $K$ in region 2+ $K$ in region 3
or, $E_{3}=(300e -100e)$$J$ $=200e$ $J$
As, $E_{2}\gt E_{3} \gt E_{1}$,
from eq. $3$, we can write
$\lambda_{2}\lt \lambda_{3} \lt \lambda_{1}$
or, $\lambda_{1}\gt \lambda_{3} \gt \lambda_{2}$
Here, $\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$ are the de Broglie wavelength in region $1$, region $2$ and region $3$ respectively.
Therefore the rank of the three regions according to the de Broglie wavelength of the electron is $\lambda_{1}\gt \lambda_{3} \gt \lambda_{2}$
