Question

When a beam of 10.6 eV photons falls on a platinum surface, 53% of incident photons eject photoelectrons. If intensity is $2.0~W/m^2$ and area is $1.0\times10^{-4}~m^2$, the number of photoelectrons emitted per second is:

• A. $6.25\times10^{13}$
• B. $11.79\times10^{13}$
• C. $62.5\times10^{11}$
• D. $11.79\times10^{11}$

Hint:

Photoelectrons = $\frac{\text{Intensity} \times \text{Area}}{E_{photon}} \times \text{Quantum Yield}$.

Answer 1

The Correct Option is A

Step 1: Concept
Total power $P = \text{Intensity} \times \text{Area}$. Number of photons $n = P / E_{photon}$.
Step 2: Calculation
$P = 2.0 \times 1.0\times10^{-4} = 2.0\times10^{-4}~W$.
$E = 10.6 \times 1.6\times10^{-19}~J$.
$n = \frac{2.0\times10^{-4}}{10.6 \times 1.6\times10^{-19}} \approx 1.179\times10^{14}$ photons/sec.
Step 3: Analysis
Photoelectrons = $n \times 53% = 1.179\times10^{14} \times 0.53 \approx 6.25\times10^{13}$.
Step 4: Conclusion
Hence, the number of photoelectrons emitted is $6.25\times10^{13}$. Final Answer:(A)