Question

A free particle with initial kinetic energy E and de-broglie wavelength $\lambda$ enters a region in which it has potential energy U. What is the particle's new de-Broglie wavelength?

• A. $\lambda (1-U/E{)}^{-1/2}$
• B. $\lambda (1-U/E)$
• C. $\lambda (1-U/E{)}^{-1}$
• D. $\lambda (1-U/E{)}^{1/2}$

Answer 1

The correct option is A $\lambda (1-U/E{)}^{-1/2}$

$\begin{array}{c}Theinitialkineticenergyoffreeparticle:\\ E=\frac{p^2}{2m}\to =\sqrt{2mE}\\ Debrogliewavelength:\\ \lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}\\ Energyoftheparticlewhenitenterstheregion:\\ E_f=E-U\\ So,itswavelengthbecomes:\\ {\lambda }_f=\frac{h}{\sqrt{2m{E}_f}}=\frac{h}{\sqrt{2m\left(E-U\right)}}\\ {\lambda }_f^2=\frac{h^2}{2mE}\left(\frac{E}{E-V}\right)=\frac{h^2}{2mE}\left(\frac{1}{1-U/E}\right)\\ {\lambda }_f=\frac{h}{\sqrt{2mE}}{\left(\frac{1}{1-U/E}\right)}^{1/2}=\lambda {\left(1-U/E\right)}^{-1/2}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.