Question

The beam of light has wavelengths \(4144\)\AA, \(4972\)\AA and \(6216\)\AA with a total intensity of \(3.6\times10^{-3}\,\text{W m}^{-2}\) equally distributed amongst the three wavelengths. The beam falls normally on an area of \(1\,\text{cm}^2\) of a clean metallic surface of work function \(2.3\,\text{eV}\). Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in \(2\,\text{s}\).

• A. \(2\times10^9\)
• B. \(1.075\times10^{12}\)
• C. \(9\times10^8\)
• D. \(3.75\times10^6\)

Hint:

Only photons with: \[ h\nu \ge \phi \] contribute to photoelectric emission.

Answer 1

The Correct Option is B

Step 1: Intensity per wavelength: \[ I = \frac{3.6\times10^{-3}}{3} = 1.2\times10^{-3}\,\text{W m}^{-2} \]
Step 2: Area \(A=1\,\text{cm}^2=10^{-4}\,\text{m}^2\). \[ P = IA = 1.2\times10^{-7}\,\text{W} \]
Step 3: Energy incident in \(2\,\text{s}\): \[ E = 2.4\times10^{-7}\,\text{J} \]
Step 4: Only wavelengths \(4144\)\AA and \(4972\)\AA have photon energy greater than work function.
Step 5: Number of photons: \[ N = \sum \frac{E}{hc/\lambda} \approx 1.075\times10^{12} \]