Question

The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75\%$ of the initial value ?

• A. $\frac{1}{9}E$
• B. $\frac{7}{9}E$
• C. $E$
• D. $\frac{16}{9}E$

Answer 1

Answer: (B)

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}},$
$mv=\sqrt{2mE}$
$\lambda \propto \frac{1}{\sqrt{E}}$
$\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{E_1}{E_2}}=\frac{3}{4},$
${\lambda }_2=0.75{\lambda }_1$
$\frac{E_1}{E_2}={\left(\frac{3}{4}\right)}^2$
$E_2=\frac{16}{9}{E}_1=\frac{16}{9}E$
$\left(E_1=E\right)$
Extra energy given $=\frac{16}{9}E-E$
$=\frac{7}{9}E$