An electron with initial kinetic energy of 100 , ,eV is accelerat… - Vidyadip

MCQ

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 $

✓
$1$
B
$\sqrt{1.5}$
C
$\sqrt{3}$
D
$12.27$

✓

Answer

Correct option: A.

$1$

a
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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