Question

If the kinetic energy of the particle is increased by $15$ times its previous value, the percentage change in the de-Broglie wavelength of the particle is ________$\%$.

Answer 1

de-Broglie wavelength is
$\lambda=\frac{ h }{ p }=\frac{ h }{\sqrt{2 mE }}$
When K.E. is increased by 15 times,
$E ^{\prime}= E +15 E =16 E$
$\therefore \lambda^{\prime}=\frac{ h }{\sqrt{2 m (16 E )}}$
$=\frac{1}{4} \frac{ h }{\sqrt{2 mE }}=\frac{\lambda}{4}=0.25 \lambda$
$\therefore \% $ change $=\lambda-\lambda^{\prime}=\lambda-0.25 \lambda$
$ =0.75 \lambda \Rightarrow 75 \%$