Question

If kinetic energy of a particle is increased by 16 times the percentage change in De broglie wavelength?

Answer 1

Dear Student,
$$\begin{gathered} \lambda =\frac{1}{\sqrt{KE}} \\ \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{{\left(KE\right)}_1}{{\left(KE\right)}_2}} \\ \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{KE}{16KE}}=\frac{1}{4} \\ {\lambda }_2=\frac{1}{4}{\lambda }_1 \\ {\lambda }_2=0.25{\lambda }_1 \\ \mathrm{Change}\mathrm{in}\mathrm{wavelength}=0.75or75\% \end{gathered}$$
Regards