The deBroglie wavelength of electron when accelerates through potential V is given by \( \lambda = \displaystyle\frac{h}{\sqrt{x}\,\sqrt{V}} \) 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
The correct answer is Option (2) → 2 me
An electron accelerated from rest through a potential difference \( V \) gains kinetic energy \( K = eV \). Its momentum follows from \( K = p^{2}/2m \), and the de Broglie wavelength is \( \lambda = h/p \).
\( \displaystyle p = \sqrt{2 m K} = \sqrt{2 m e V} \)
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e}\,\sqrt{V}} \)
Comparing with \( \lambda = \frac{h}{\sqrt{x}\,\sqrt{V}} \) gives \( x = 2me \).