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Question

(D) The initial kinetic energy of the particle is $K_i = E$.The initial de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK_

Vedclass · Accepted Answer

$\lambda \left( 1 - \frac{V}{E} \right)^{-\frac{1}{2}}$

Vedclass · Answer

(D) The initial kinetic energy of the particle is $K_i = E$.The initial de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK_i}} = \frac{h}{\sqrt{2mE}}$.When the particle enters a region with potential energy $V$,its total energy remains conserved. The new kinetic energy $K_f$ is given by $K_f = E - V$.The new de-Broglie wavelength $\lambda'$ is $\lambda' = \frac{h}{\sqrt{2mK_f}} = \frac{h}{\sqrt{2m(E - V)}}$.We can rewrite this as $\lambda' = \frac{h}{\sqrt{2mE(1 - \frac{V}{E})}} = \frac{h}{\sqrt{2mE}} \cdot \frac{1}{\sqrt{1 - \frac{V}{E}}}$.Substituting $\lambda = \frac{h}{\sqrt{2mE}}$,we get $\lambda' = \lambda \left( 1 - \frac{V}{E} \right)^{-\frac{1}{2}}$.

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

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