Question

If the velocity of a particle is increased three times then the percentage decreases in its de Broglie wavelength will be

Answer 1

$\lambda =\frac{1}{\sqrt{KE}}=\frac{1}{\sqrt{\frac{1}{2}m{v}^2}}\propto \frac{1}{v}$

Given $v_2=3v_1$

$\Rightarrow {\lambda }_1{v}_1={\lambda }_2{v}_2$

$\Rightarrow {\lambda }_1{v}_1={\lambda }_2\times \left(3v_1\right)$

$\Rightarrow {\lambda }_1{v}_1=3{\lambda }_2{v}_1$

$\Rightarrow {\lambda }_1=3{\lambda }_2$

$\Rightarrow {\lambda }_2=\frac{1}{3}{\lambda }_1$

Percent $\Rightarrow \frac{1}{3}\times 100=33.33$%