A small metal plate (work-function W_o) is kept at a distance d from a singly ionised fixed ion. A monochromatic light beam is incident on the metal plate and photoelectric are emitted. The maximum wavelength of light so that the photo-electrons may go round the ions along a circle is

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$\frac{8 {πε}_{0} W_{0} d + e^{2}}{8 {πhcε}_{0} d}$

$\frac{8 {πε}_{0} d}{{hceW}_{0}}$

$\frac{8 {πhcε}_{0} d}{8 {πε}_{0} W_{0} d + e^{2}}$

$\frac{2 (\frac{hc}{W_{0}} - e^{2})}{n}$

answer is B.

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Detailed Solution

Here we have
$\frac{{mv}^{2}}{d} = \frac{e^{2}}{4 {πε}_{0} d^{2}} or v^{2} = \frac{e^{2}}{4 {πε}_{0} md} . . . . . (1) By Einstein' s photoelectric equation h ν = W_{0} + \frac{1}{2} {mv}^{2} = W_{0} + \frac{1}{2} {mv}^{2} or, \frac{2 (hc / λ - W_{0})}{m} = v^{2} . . . . . (2) From eqs . (1) and (2), we have \frac{2 (\frac{hc}{λ} - W_{0})}{m} = \frac{e^{2}}{4 {πε}_{0} md} or \frac{hc}{λ} = W_{0} + \frac{e^{2}}{8 {πε}_{0} d} or \frac{hc}{λ} = \frac{8 {πε}_{0} W_{0} d + e^{2}}{8 {πε}_{0} d} or λ = \frac{8 {πε}_{0} hcd}{8 {πε}_{0} W_{0} d + e^{2}}$