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

Answer: 75

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