Question

The potential energy of a particle of mass m is given by
$U\left(x\right)=\{\begin{array}{cc}{E}_0;& 0\le x\le 1\\ 0;& x>1\end{array}$
${\lambda }_1$ and ${\lambda }_2$ are the de-Broglie wavelengths of the particle, when $0\le x\le 1$ and $x>1$ respectively. If the total energy of particle is $3E_0$, the ratio $\frac{{\lambda }_1}{{\lambda }_2}$ will be

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

Answer 1

The correct option is D $\sqrt{\frac{3}{2}}$

$K.E=3E_0-E_0=2E_0(for0\le x\le 1)\Rightarrow {\lambda }_1=\frac{h}{\sqrt{2m_2{E}_0}}$
$K.E.=3E_0(forx>1)\Rightarrow {\lambda }_2=\frac{h}{\sqrt{2m_3{E}_0}}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{3}{2}}$