Question

The beam of light has three wavelengths 4144AA, 4972AA, 6216AA with a total intensity of 3.6×10⁻3W/m² equally distributed among the three wavelengths. The beam falls normally on a clean metallic surface of work function 2.3eV. Assuming no loss by reflection and that each photon ejects one photoelectron, calculate the number of photoelectrons liberated in 2s.

• A. \( 2\times10^9 \)
• B. \( 1.075\times10^{12} \)
• C. \( 9\times10^8 \)
• D. 3.75×10⁶

Hint:

Always check threshold wavelength before counting photoelectrons.

Answer 1

The Correct Option is B

Step 1: Power per wavelength: P = frac3.6×10⁻33 = 1.2×10⁻3W
Step 2: Only wavelengths with photon energy greater than work function eject electrons. Shorter wavelengths satisfy this.
Step 3: Number of photons per second: N = (P)/(hc/λ)
Step 4: Total photoelectrons in 2s: N = 1.075×10¹2