The deBroglie wavelength of electron when accelerates through potential V is given by $\lambda=\frac{h}{\sqrt{x} \sqrt{V}}$ then x is equal to :
where m = mass of electron
k = kinetic energy of electron
p = momentum
e = charge
Answer & explanation
Correct answer: option 2
Debroglie wavelength is given by
$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2mqV}} = \frac{h}{\sqrt{2meV}}$
$\Rightarrow x = 2me$