The Kinetic Energy of an electron having de Broglie wavelength($\lambda$) triple , then the de Broglie wavelength associated with it becomes:
Answer & explanation
Correct answer: option 3
$\lambda = \frac{h}{\sqrt {2mE}}$
$\Rightarrow \lambda \propto \frac{1}{\sqrt E}$
$\lambda'= \frac{\lambda}{\sqrt 3}$