The potential energy of a particle of mass $m$ is given by $U(x) = \left\{ \begin{array}{r} E_{0};\ 0 \leq x \leq 1 \ 0;\ \ \ \ \ \ \ x 1 \end{array} ight.\ $ $\lambda_{1}$ and $\lambda_{2}$ are the de-Broglie wavelengths of the particle, when $0 \leq x \leq 1$ and $x 1$ respectively. If the total energy of particle is $2E_{0},$ the ratio $\frac{\lambda_{1}}{\lambda_{2}}$ will be

Question

The potential energy of a particle of mass $m$ is given by $U(x) = \left\{ \begin{array}{r} E_{0};\ 0 \leq x \leq 1 \ 0;\ \ \ \ \ \ \ x > 1 \end{array} ight.\ $ $\lambda_{1}$ and $\lambda_{2}$ are the de-Broglie wavelengths of the particle, when $0 \leq x \leq 1$ and $x > 1$ respectively. If the total energy of particle is $2E_{0},$ the ratio $\frac{\lambda_{1}}{\lambda_{2}}$ will be

Dr. Abubacker · Accepted Answer

(c) $K.E. = 2\ E_{0} - E_{0} = E_{0}\ (for\ 0 \leq x \leq 1) \Rightarrow \lambda_{1} = \frac{h}{\sqrt{2mE_{0}}}$
$K.E.\  = 2E_{0}(for\ x > 1) \Rightarrow \lambda_{2} = \frac{h}{\sqrt{4mE_{0}}} \Rightarrow \frac{\lambda_{1}}{\lambda_{2}} = \sqrt{2}$

Explanation

Reviewed by Dr. Abubacker

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