Question

The potential energy of particle of mass $m$ varies as
$U(x)=\{\begin{array}{ll}{E}_0& \text{for 0≤x≤1 }\\ 0& \text{for x>1 }\end{array}$
The de-Broglie wavelength of the particle in the range $0\le x\le 1$ is ${\lambda }_1$ and that in the range $x>1$ is ${\lambda }_2$ . If the total energy of the particle is $2E_0$ , find $\frac{{\lambda }_1}{{\lambda }_2}$ .

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{\frac{1}{2}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

Answer: (A)

We know that
$\lambda =\frac{h}{mv}$
$\lambda =\frac{h}{\sqrt{2mk}}$
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{k_2}{k_1}}$ ($\because h$ and $m$ are constants)
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{2k}{k}}=\sqrt{2}$