91Ó°ÊÓ

Kinetic energy is the difference between total energy and rest energy.

$K=\frac{1}{2}m{\mathrm{Ï\dots }}^2$

The momentum is the product of the mass of a particle and the speed of a particle.

$p=m\mathrm{Ï\dots }$

The de Broglie wavelength of a particle is inversely proportional to the momentum of a particle.

$\mathrm{λ}=\frac{h}{p}$

Given that, the electron has energy$E=3U_0-U$

The wavelength for$0<x<L$

If the electron is in the region$0<x<L$, the potential energy is$U=0$. So,

Kinetic energylocalid="1663998094549" $K=3U_0$

$$\begin{gathered} {\mathrm{λ}}_{\left(0<x<L\right)}=\frac{h}{\sqrt{2mK}} \\ ==\frac{h}{\sqrt{6m{U}_0}} \end{gathered}$$

The wavelength for $x>L$

If the electron is in the region $x>L$, the potential energy is $U=U_0$. So, kinetic energy is

$K=3U_0-U_0$

$=2U_0$

$$\begin{gathered} {\mathrm{λ}}_{\left(x<L\right)}=\frac{h}{\sqrt{2mK}} \\ ==\frac{h}{\sqrt{4m{U}_0}} \end{gathered}$$

Now, the ratio of wavelengths:

$$\begin{gathered} \frac{{\mathrm{λ}}_{\left(x>L\right)}}{{\mathrm{λ}}_{\left(0<x>L\right)}}=\frac{h/\sqrt{4m{U}_0}}{h/\sqrt{6m{U}_0}} \\ =\sqrt{\frac{3}{2}} \end{gathered}$$

Hence, the ratio of the de Broglie wavelength of the electron in the region $x>L$to the wavelength for$0<x>L$ is $\sqrt{3/2}$.