Question

The ratio of the power of a light source \( S_1 \) to that of the light source \( S_2 \) is 2. \( S_1 \) is emitting \( 2 \times 10^{15} \) photons per second at 600 nm. If the wavelength of the source \( S_2 \) is 300 nm, then the number of photons per second emitted by \( S_2 \) is ________________ \( \times 10^{14} \).

Hint:

The number of photons emitted is inversely proportional to the energy of each photon. For shorter wavelengths, the energy per photon is higher, so fewer photons are emitted for the same power.

Answer 1

Step 1: Input Data Analysis

The following parameters are provided:

• Power ratio of light sources \( S_1 \) and \( S_2 \): \( \frac{P_1}{P_2} = 2 \).
  
• Photon emission rate of source \( S_1 \): \( 2 \times 10^{15} \) photons/second at a wavelength of 600 nm.
  
• Wavelength of source \( S_2 \): 300 nm.
  

Step 2: Fundamental Relationships

Light source power is defined by the energy radiated per unit time. The energy of a single photon is determined by its wavelength using the formula:
E = \frac{hc}{\lambda}
where:
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \) J·s),
- \( c \) = speed of light (\( 3.0 \times 10^{8} \) m/s),
- \( \lambda \) = wavelength of the light.
The power output of a source is the product of the number of photons emitted per second and the energy of each photon:
P = Number of photons × E
Consequently, the number of photons emitted per second is directly proportional to the power and inversely proportional to the energy per photon.

Step 3: Power and Photon Emission Correlation

The given power ratio \( \frac{P_1}{P_2} = 2 \) can be expressed as:
P₁ / P₂ = (N₁ × E₁) / (N₂ × E₂) = 2
where:
- N₁ and N₂ denote the photon emission rates (photons per second) for \( S_1 \) and \( S_2 \), respectively.
- E₁ and E₂ represent the energy per photon for \( S_1 \) and \( S_2 \), respectively.
The energy per photon is inversely proportional to the wavelength, leading to:
E₁ / E₂ = λ₂ / λ₁
Given λ₁ = 600 nm and λ₂ = 300 nm, the energy ratio is:
E₁ / E₂ = 300 / 600 = 1 / 2
Substituting this energy ratio into the power equation yields:
(N₁ × (1/2)) / N₂ = 2
Simplifying this equation results in:
N₁ / (2 N₂) = 2 \implies N₁ = 4 N₂

Step 4: Photon Emission Rate Calculation for \( S_2 \)

Using the given photon emission rate for \( S_1 \) (\( N_1 = 2 \times 10^{15} \) photons/second) and the derived relationship \( N_1 = 4 N_2 \):
2 \times 10^{15} = 4 N₂ \implies N₂ = (2 \times 10^{15}) / 4 = 0.5 \times 10^{15}
Therefore, the photon emission rate for \( S_2 \) is \( 5 \times 10^{14} \) photons/second.

Conclusion

The photon emission rate of source \( S_2 \) is calculated to be 5 × 10¹⁴ photons per second.