Question

A small metal plate of work function ${}^{\prime }{\varphi }^{\prime }$ is kept at a distance ${}^{\prime }{d}^{\prime }$ from a singly ionised fixed ion. A monochromatic light beam incidents on the metal plate and photoelectrons are emitted. The maximum wavelength of the light beam, so that some of the electrons may go around the ion along a circle, is

• A. $\frac{4\pi {\epsilon }_{0}hcd}{e^2+8\pi {\epsilon }_0\varphi d}$
• B. $\frac{8\pi {\epsilon }_{0}hcd}{e^2+8\pi {\epsilon }_0\varphi d}$
• C. $\frac{hc}{8\pi {\epsilon }_0\varphi }$
• D. $\frac{hc}{4\pi {\epsilon }_0\varphi }$

Answer 1

The correct option is B $\frac{8\pi {\epsilon }_{0}hcd}{e^2+8\pi {\epsilon }_0\varphi d}$

For metal plate work function $\varphi$

Charge $q=+e$,

Wavelength of light beam = $\lambda$

Energy of photo electron = $K{E}_{min}=\frac{hC}{\lambda }-\varphi =\frac{1}{2}{m}_e{V}^2$ --- (1)

An electron moving with velocity V will require a force of $\frac{m{V}^2}{R}$ perpendicular to its velocity. Here, the maximum length of radius is d.

$\therefore \frac{m_e{V}^2}{d}=$centripetal force --- (2)

This force will be provided by electrostatic field of ion.

$F=\frac{1}{4\pi E_0}\frac{e^2}{R^2}[hereR=d]$ --- (3)

$\therefore$ equating them, using (1), (2) and (3)

$\frac{m_e{V}^2}{d}=\frac{1}{4\pi E_0}\frac{e^2}{d^2}$

$\Rightarrow \frac{2(hC-\varphi \lambda )}{\lambda d}=\frac{1}{4\pi E_0}\frac{e^2}{d^2}$

$\Rightarrow 8\pi E_{0}dhC-8\varphi \lambda \pi E_{0}d=e^2\lambda$

$\therefore \lambda =\frac{8\pi E_{0}dhC}{e^2+8\pi E_{0}d\varphi }$