Question
An electron with initial kinetic energy of $100 \,\,eV$ is acceleration through a potential difference of $50\, V$. Now the de-Broglie wavelength of electron becomes .................. $\mathop {\rm{A}}\limits^o $
✓
Answer
Additional kinetic energy it gains is
$K^{\prime}=e V=50 \quad e V$
Total kinetic energy is equal to $150 \times 1.6 \times 10^{-19} J$
The de broglie wavelength is $\lambda=\frac{h}{\sqrt{2 m K}}=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-34} \times 150 \times 1.6 \times 10^{-19}}}=1 \times 10^{-10} \mathrm{m}$
$\lambda=1 A^{o}$
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