The kinetic Energy of an electron having de Broglie wavelength triple, then the de Broglie wavelength associated with it becomes:
Answer & explanation
Correct answer: option 3
$ \lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mE}} $
$\lambda \propto \frac{1}{\sqrt E}$
$ \lambda \text{ will become }\frac{1}{\sqrt 3 } \text{ times when E is increased to 3 times.}$