Question
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 ?
✓
Answer
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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