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:
Answer & explanation
Correct answer: option 2
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,
$λ'=\frac{h}{\sqrt{2me9V}}=\frac{4}{3\sqrt{2meV}}=\frac{λ}{3}$