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Q.
The de-Broglie wavelength of an electron accelerated through a potential difference of $100 \, V$ is
NTA AbhyasNTA Abhyas 2022
Solution:
De-Broglie wavelength of an electron is given by
$\lambda =\frac{h}{m v}=\frac{h}{\sqrt{2 m e V}}-\frac{12.27}{\sqrt{V}}Å$
( $h=$ Planck's constant)
Where, $m=$ mass of electron, $e=$ electronic charge and $V=$ potential difference with which electron is accelerated.
$\lambda =\frac{12.27}{\sqrt{100}}Å=\frac{12.27}{10}Å$
$=1.227 \, Å$