Question

In photoelectric effect, initially when energy of electrons emitted is $E_0$, de-Broglie wavelength associated with them is ${\lambda }_0$. Now, energy is doubled then associated de-Broglie wavelength ${\lambda }^{{}^{\prime }}$ is

• A. ${\lambda }^{{}^{\prime }}=\frac{{\lambda }_0}{\sqrt{2}}$
• B. ${\lambda }^{{}^{\prime }}=\sqrt{2}{\lambda }_0$
• C. ${\lambda }^{{}^{\prime }}={\lambda }_0$
• D. ${\lambda }^{{}^{\prime }}=\frac{{\lambda }_0}{2}$

Answer 1

The correct option is A ${\lambda }^{{}^{\prime }}=\frac{{\lambda }_0}{\sqrt{2}}$

de-Brogile wavelength is given by
$\lambda =\frac{h}{p}$, where h= Planck's constant and p= momentum
Also energy (E) and momentum are related as
$E=\frac{p^2}{2m}$
$\therefore p=\sqrt{2mE}$
$\therefore \lambda =\frac{h}{\sqrt{2mE}}\times \frac{1}{\sqrt{E}}$ as h and m are constants
Hence, $\frac{{\lambda }_0}{{\lambda }^{\prime }}=\sqrt{\frac{E^{\prime }}{E}}=\sqrt{\frac{2E}{E}}=\sqrt{2}$
$\therefore {\lambda }^{\prime }=\frac{{\lambda }_0}{\sqrt{2}}$