Question

The potential energy of particle of mass m varies as $U\left(x\right)=\{\begin{array}{c}{E}_0:0\le x\le 1\\ 0:x>1\end{array}$ The de Broglie wavelength of the particle in the range $0\le x\le 1is{\lambda }_1$ and that in the range $x>1is{\lambda }_2$. If the total energy of the particle is $2E_0$, find ${\lambda }_1,/{\lambda }_2.$

• A. $\sqrt{2}$
• B. $2$
• C. $1$
• D. $\frac{2}{\sqrt{2}}$

Answer 1

The correct option is A $\sqrt{2}$

de Broglie wavelength of a particle of ///

$\lambda =\frac{h}{\sqrt{2mK}}$

where K is the kinetic energy of the particle.

For $0\le x\le 1,U\left(x\right)=E_0$

As, total energy = Kinetic energy + potential energy

$\therefore 2E_0=K_1+E_0$

or $K_1=E_0$

$\therefore {\lambda }_1=\frac{h}{\sqrt{2m{K}_1}}=\frac{h}{\sqrt{2m{E}_0}}$

For $x>1,U\left(x\right)=0\therefore K_2=2E_0$

$\therefore {\lambda }_2=\frac{h}{\sqrt{2m{K}_2}}=\frac{h}{\sqrt{2m\left(2E_0\right)}}$

Divide (i) by (ii), we get

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{h}{\sqrt{2m{E}_0}}\times \frac{\sqrt{2m\left(2E_0\right)}}{h}=\sqrt{2}$