Question

In the photoelectric experiment, if we use a monochromatic light, the photoelectric current vs voltage curve is as shown. If work function of the metal is 2eV, estimate the power of light used. (Assume efficiency of photo emission = ${10}^{-3}\%$, i.e., number of photoelectrons emitted are ${10}^{-3}\%$ of number of photons incident on metal)

• A. 2W
• B. 5W
• C. 7W
• D. 10W

Answer 1

Answer: (C)

Step 1: Given data -

The work function of the metal is ${\phi }_0=2eV$

The efficiency of the photon emission is $e={10}^{-3}\%$

From the given diagram,

The stopping potential (the value of potential at which the current is zero) is $V_s=5V$

The saturation current (the peak value of the photoelectric current) is $I_s=10\mu A$

Step 2: Formula used -

According to the photoelectric equation, the energy of the incident ray is

$hv=e{V}_s+{\phi }_0$

Where h is Plank’s constant, is the frequency of the incident ray, and e is the fundamental value of charge,

Power of the light in terms of its energy is,

$p=\frac{n_{ph}hv}{t}$

Where t is the time taken and $n_{ph}$ is the number of the incident photons,

The efficiency in terms of the number of incident photons and emitted electrons is,

$e=\frac{n_e}{n_{ph}}\times 100$

Where ne is the number of emitted electrons,

The photoelectric current in terms of the number of electrons emitted is,

$I=\frac{n_{e}e}{t}$

Where t is the time taken,

Step 3: Calculating the power of light -

By substituting the known values in the photoelectric equation, we get,

$hv=5e+2e$

$hv=7e$

By substituting the known values in equation of efficiency, we get,

${10}^{-3}\%=\frac{n_e}{n_{ph}}\times 100\%$

$\frac{{10}^{-3}}{100}=\frac{n_e}{n_{ph}}$

${10}^{-5}=\frac{n_e}{n_{ph}}$

$n_e={10}^{-5}{n}_{ph}$

Thus, the value of photoelectric current is,

$I=\frac{{10}^{-5}{n}_{ph}e}{t}$

Or

$t=\frac{{10}^{-5}{n}_{ph}e}{I}$

By using the equation (3), the power of the light becomes,

$P=\frac{n_{ph}hv}{\frac{{10}^{-5}{n}_{ph}e}{I}}$

$p=n_{p}hv\times \frac{I}{{{10}^{}}^{-5}{n}_{ph}e}$

$P=\frac{Ihv}{{10}^{-5}e}$

By substituting the value of saturation current and the value of energy from equation (1),

$P=\frac{10\times {10}^{-6}\times 7e}{{10}^{-5}e}$

$P=\frac{{10}^{-5}\times 7e}{{10}^{-5}e}$

$P=7W$

Thus, the power of the incident light is 7 W.

Hence, option C is the correct answer.