A free particle with initial kinetic energy E and de-broglie wave… - Vidyadip

MCQ

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 ?

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

✓

Answer

Correct option: D.

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

d
Initial kinetic energy.

$\mathrm{K}_{\mathrm{i}}=\mathrm{E}$

de-broglie wavelength

$\lambda = \frac{{\rm{h}}}{{\rm{p}}} = \frac{{\rm{h}}}{{\sqrt {2{\rm{m}}{{\rm{K}}_i}} }} = \frac{{\rm{h}}}{{\sqrt {2{\rm{mE}}} }}$

Final kinetic energy:

$\mathrm{K}_{\mathrm{f}}=\mathrm{E}-\mathrm{V}$

$\lambda^{\prime}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{f}}}}$

$\lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}(\mathrm{E}-\mathrm{V})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}} \sqrt{1-\frac{\mathrm{V}}{\mathrm{E}}}}$

$\lambda^{\prime}=\frac{\lambda}{\sqrt{1-\frac{V}{E}}}$

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