The potential energy of a particle of mass $m$ is given by $$ V(x)=\left\{\begin{array}{cl} E_{0} ;

Question

The potential energy of a particle of mass $m$ is given by $$ V(x)=\left\{\begin{array}{cl} E_{0} ; & \leq x \leq 1 \ 0 ; & x>1 \end{array}ight\} $$ $\lambda_{1}$ 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 $2 E_{0}$, find $\lambda_{1}^2 / \lambda_{2}^2$.\

Professor PC Thomas Classes · Answer

de Broglie wavelength $\lambda=\frac{h}{p}$\ or $\lambda=\frac{h}{\sqrt{2 m K}}$\ where $K$ denotes kinetic energy of particle.\ Case (i) : $0 \leq x \leq 1$\ Given: Potential energy $=E_{0}$\ Given: Total energy $=2 E_{0}$\ $	herefore \quad$ Kinetic energy $=2 E_{0}-E_{0}=E_{0}$\ $	herefore \lambda_{1}=\frac{h}{\sqrt{2 m E_{0}}}$\ Case (ii): $x>1$\ Given: Potential energy $V(x)=0$\ Given: $\quad$ Total energy $=2 E_{0}$\ $	herefore \quad$ Kinetic energy $=2 E_{0}$\ $	herefore \quad \lambda_{2}=\frac{h}{\sqrt{2 m\left(2 E_{0}ight)}} \quad 	herefore \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{2 E_{0}}{E_{0}}}=\sqrt{2}$\ $	herefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{2}$.\