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Question

Ultraviolet light of wavelength 300 nm and intensity 1.0 ${\rm{W}}{{\rm{m}}^{ - 2}}$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 ${\rm{c}}{{\rm{m}}^2}$ of the surface is nearly

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
**Primary Concept**: Photoelectric effect and relationship between intensity, photon energy, and number of photons.
**Related Topics**: Electromagnetic radiation, Planck's constant, energy of a photon, conversion of units.
**Prerequisites**: Understanding of photon energy, intensity, and the photoelectric effect.
## SOLUTION PATHWAY
**Given Information**:
* Wavelength (λ) = 300 nm = 300 x 10^-9 m
* Intensity (I) = 1.0 W/m^2
* Area (A) = 1.0 cm^2 = 1.0 x 10^-4 m^2
* Efficiency = 1% = 0.01
**Method & Steps**:
1. Calculate the energy of a single photon (E) using the formula E = hc/λ, where h is Planck's constant (6.626 x 10^-34 J.s) and c is the speed of light (3 x 10^8 m/s).
   Reasoning: To find the energy carried by each photon.
   Formula/Rule: E = hc/λ
2. Calculate the power incident on the area by using P = I*A
   Working: P = 1 W/m^2 * 1 x 10^-4 m^2 = 1 x 10^-4 W
   Key Point: Power is energy per unit time.
3. Calculate the number of photons incident per second by dividing the total power by the energy of one photon. N_photons = P/E
   Working: E = (6.626 x 10^-34 J.s * 3 x 10^8 m/s) / (300 x 10^-9 m) = 6.626 x 10^-19 J. N_photons = (1 x 10^-4 W) / (6.626 x 10^-19 J) = 1.509 x 10^14 photons/s
   Key Point:  Intensity is power per unit area.
4. Calculate the number of photoelectrons emitted per second by multiplying the number of incident photons by the efficiency. N_electrons = N_photons * efficiency
   Working: N_electrons = 1.509 x 10^14 photons/s * 0.01 = 1.509 x 10^12 electrons/s
   Result: Approximately 1.51 x 10^12 electrons/s
   Verification: The calculated value matches the correct answer.
## OPTION ANALYSIS
[A] :x: This value is incorrect because it doesn't take into account the 1% efficiency of photoelectron emission. It represents the number of incident photons, not photoelectrons.
[B] :x: This value is incorrect. It's lower than the correct answer, likely due to a miscalculation of the number of photons or the application of the efficiency.
[C] ✓: This is the correct answer. The calculation of the number of photons, and the application of the 1% efficiency is correctly done.
[D] :x: This value is too low, indicating a significant error in the calculations or conceptual understanding.
## LEARNING AIDS
**Common Mistakes**:
1. **Using the wrong units**:  Ensure all units are in SI units (meters, seconds, Joules, etc.) before calculations.  Convert cm^2 to m^2.
2. **Forgetting the efficiency**:  Remember to multiply the number of incident photons by the given efficiency (1%) to find the number of photoelectrons.
**Quick Tips**:
1. **Photon energy formula**:  Memorize the formula E = hc/λ and how to use it.
2. **Intensity and power**: Remember that intensity is power per unit area (I = P/A), so power is intensity times area (P=IA).

SUBHANI · Answer

$9.61 	imes {10^{14}}{{m{s}}^{ - 1}}$

SUBHANI · Answer

$4.12 	imes {10^{13}}{{m{s}}^{ - 1}}$

SUBHANI · Answer

The correct Option is "$1.51 	imes {10^{12}}{{m{s}}^{ - 1}}$"

SUBHANI · Answer

$2.13 	imes {10^{11}}{{m{s}}^{ - 1}}$

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