In the photoelectric experiment, if we use a monochromatic light, the $$I\text{-}V$$ curve is as shown. If the work function of the metal is $$2\ \mathrm{eV}$$, estimate the power of light used.

(Assume efficiency of photo emission = $$10^{-3}\%$$, i.e. the number of photoelectrons emitted are $$10^{-3}\%$$ of the number of photons incident on the metal.)

The stopping potential gives the maximum kinetic energy of the photoelectrons:  $$K_{\max}=eV_0$$

From the graph,  $$V_0=5\,V$$

Hence,  $$K_{\max}=5\,eV$$

Using Einstein's photoelectric equation,  $$h\nu=\phi+K_{\max}$$

Given,  $$\phi=2\,eV$$

Therefore,  $$h\nu=2+5=7\,eV$$

Now, the saturation current is due to the photoelectrons emitted per second.

If $$n$$ is the number of incident photons per second, then due to the efficiency,  $$n_e=10^{-5}n$$

Therefore,  $$I_s=en_e=e(10^{-5}n)$$

Given,  $$I_s=10\,\mu A=10^{-5}\,A$$

Thus,  $$10^{-5}=e(10^{-5})n$$

$$n=\frac{1}{e}$$

The power of incident light is $$P=nh\nu$$

Since, $$h\nu=7\,eV=7e\,J$$

we get  $$P=\frac{1}{e}\times7e$$

$$P=7\,W$$

Therefore,  $${P=7\,W}$$

Hence, the correct option is B.