The de-Broglie wavelength associated with an electron accelerated through a potential difference V is λ. The new de Broglie wavelength of electron when potential difference is increased to 9V will be:

$\frac{λ}{3}$

The correct answer is Option (2) → $\frac{λ}{3}$

According to De-broglie,

$λ=\frac{h}{P}=\frac{h}{\sqrt{2meV}}$

$∴λ∝\frac{1}{\sqrt{V}}$

and,

$λ'=\frac{h}{\sqrt{2me9V}}=\frac{4}{3\sqrt{2meV}}=\frac{λ}{3}$