A 600 W mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second?

Answer

To find the number of photons emitted from a 600 W mercury lamp that emits monochromatic radiation of wavelength 331.3 nm, we can follow these steps:

### Step 1: Convert the wavelength from nanometers to meters
The wavelength given is 331.3 nm. To convert this to meters, we use the conversion factor \(1 \, \text{nm} = 10^{-9} \, \text{m}\).

\[ \text{Wavelength} \, (\lambda) = 331.3 \, \text{nm} = 331.3 \times 10^{-9} \, \text{m} = 3.313 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate the energy of one photon
The energy of a photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \, \text{J s}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda\) = \(3.313 \times 10^{-7} \, \text{m}\)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{3.313 \times 10^{-7} \, \text{m}} \]

Calculating this gives:

\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{3.313 \times 10^{-7} \, \text{m}} \approx 5.99 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the total energy emitted per second
The lamp emits 600 W of power, which means it emits 600 joules of energy per second.

\[ \text{Power} = 600 \, \text{W} = 600 \, \text{J/s} \]

### Step 4: Calculate the number of photons emitted per second
To find the number of photons emitted per second, we can use the formula:

\[ \text{Number of photons} = \frac{\text{Total energy per second}}{\text{Energy per photon}} \]

Substituting the values we have:

\[ \text{Number of photons} = \frac{600 \, \text{J/s}}{5.99 \times 10^{-19} \, \text{J}} \]

Calculating this gives:

\[ \text{Number of photons} \approx 1.00 \times 10^{21} \, \text{photons/s} \]

### Final Answer
The number of photons emitted from the lamp per second is approximately \(1.00 \times 10^{21}\) photons/s.

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