A modern 200 W sodium street lamp emits yellow light of wavelength 0.6 `mum`. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

Options

A. `62xx10^(20)`
B. `3xx10^(19)`
C. `1.5xx10^(20)`
D. `6xx10^(18)`

Explanation

To solve the problem, we need to find the number of photons emitted per second by a sodium street lamp that has a power of 200 W and is 25% efficient in converting electrical energy to light. The wavelength of the emitted yellow light is 0.6 µm.

### Step-by-Step Solution:

1. **Calculate the Energy of One Photon:**
The energy of a photon can be calculated using the formula:
\[ E = \frac{12400}{\lambda} \]
where \( \lambda \) is the wavelength in Angstroms (Å). First, we convert the wavelength from micrometers (µm) to Angstroms (Å):
\[ 0.6 \, \mu m = 0.6 \times 10^4 \, Å = 6000 \, Å \]
Now, substituting the value of \( \lambda \) into the energy formula:
\[ E = \frac{12400}{6000} = 2.0667 \, eV \]

2. **Calculate the Effective Power Emitted as Light:**
The lamp is 25% efficient, so the effective power emitted as light can be calculated as:
\[ P_{\text{light}} = 0.25 \times 200 \, W = 50 \, W \]

3. **Convert Power from Watts to Electron Volts per Second:**
Since 1 Watt = 1 Joule/second and 1 Joule = \( 6.242 \times 10^{12} \, eV \), we convert 50 W to electron volts per second:
\[ P_{\text{light}} = 50 \, W = 50 \times 6.242 \times 10^{12} \, eV/s = 3.121 \times 10^{14} \, eV/s \]

4. **Calculate the Number of Photons Emitted per Second:**
The number of photons emitted per second can be calculated by dividing the total energy emitted per second by the energy of one photon:
\[ n = \frac{P_{\text{light}}}{E} = \frac{3.121 \times 10^{14} \, eV/s}{2.0667 \, eV} \approx 1.51 \times 10^{14} \, \text{photons/s} \]

### Final Answer:
The number of photons emitted per second by the sodium street lamp is approximately:
\[ n \approx 1.51 \times 10^{14} \, \text{photons/s} \]