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

2 p 2 me 1.2 me 2 mk

2 me

Debroglie wavelength is given by  $ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2mqV}} = \frac{h}{\sqrt{2meV}}$ $\Rightarrow x = 2me$