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)

$\lambda=\frac{ h }{ mv }=\frac{ h }{\sqrt{2 mE }},$
$ mv =\sqrt{2 mE }$
$\lambda \propto \frac{1}{\sqrt{ E }}$
$\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 $
$=\frac{7}{9} E$