Question

The potential energy of a particle varies as
$V\left(x\right)=E_o;0\le x\le 1$
$=0;x>1$
For $0\le x\le 1$, de Broglie wavelength is ${\gamma }_1$ and for $x>1$ the de Broglie wavelength is ${\gamma }_2$.

Total energy of the particle is $2E_o$. If ${\gamma }_1/{\gamma }_2=\sqrt{x}$. Find $x$

Answer 1

For $0\le x\le 1$, Kinetic energy $E_1=E_0=\frac{p_1^2}{2m}\Rightarrow p_1=\sqrt{2m{E}_0}$
For $x>1$, $E_2=2E_0=\frac{p_2^2}{2m}\Rightarrow p_2=\sqrt{4m{E}_0}$
now, ${\gamma }_1=\frac{h}{p_1}=\frac{h}{\sqrt{2m{E}_0}}$ and ${\gamma }_2=\frac{h}{p_2}=\frac{h}{\sqrt{4m{E}_0}}$
$\therefore \frac{{\gamma }_1}{{\gamma }_2}=\frac{2}{\sqrt{2}}=\sqrt{2}$