A 600 W mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second?
`(h =6.626 xx10^(-34)" Js, velocity of light "=3xx10^(8)ms^(-1))`

Options

A. `1xx10^(21)`
B. `1xx10^(23)`
C. `1xx10^(22)`
D. `1xx10^(20)`

Answer

To solve the problem of how many photons are emitted from a 600 W mercury lamp per second, we can follow these steps:

### Step 1: Understand the relationship between power, energy, and photons
The power of the lamp (P) is given as 600 W, which means it emits 600 joules of energy per second. The energy of a single photon (E) can be calculated using the formula:

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

where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^{8} \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the radiation (\(331.3 \, \text{nm} = 331.3 \times 10^{-9} \, \text{m}\)).

### Step 2: Calculate the energy of a single photon
Substituting the values into the formula:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^{8} \, \text{m/s})}{331.3 \times 10^{-9} \, \text{m}} \]

Calculating the numerator:

\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{Jm} \]

Now, divide by the wavelength:

\[ E = \frac{1.9878 \times 10^{-25}}{331.3 \times 10^{-9}} \approx 5.99 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the number of photons emitted per second
Now we can find the number of photons emitted per second (N) using the formula:

\[ N = \frac{P}{E} \]

Substituting the values:

\[ N = \frac{600 \, \text{J/s}}{5.99 \times 10^{-19} \, \text{J}} \approx 1.00334 \times 10^{21} \]

### Step 4: Round off the answer
Rounding off, we find:

\[ N \approx 1.0 \times 10^{21} \, \text{photons/s} \]

### Final Answer
The number of photons emitted from the lamp per second is approximately \(1.0 \times 10^{21}\).

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