The number of photons emitted per second by a `62W` source of monochromatic light of wavelength `4800 A^(@)` is

Options

A. `1.5xx10^(19)`
B. `1.5xx10^(20)`
C. `2.5xx10^(20)`
D. `4xx10^(20)`

Explanation

To find the number of photons emitted per second by a 62 W source of monochromatic light with a wavelength of 4800 Å, we can follow these steps:

### Step 1: Understand the relationship between power, energy, and number of photons
The power (P) of the light source is the energy emitted 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 (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light in meters.

### Step 2: Convert the wavelength from Ångstroms to meters
Given \( \lambda = 4800 \, \text{Å} \):
\[ \lambda = 4800 \times 10^{-10} \, \text{m} = 4.8 \times 10^{-7} \, \text{m} \]

### Step 3: Calculate the energy of one photon
Now, substitute the values into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{4.8 \times 10^{-7} \, \text{m}} \]

Calculating the numerator:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{4.8 \times 10^{-7} \, \text{m}} \]
\[ E \approx 4.14 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the number of photons emitted per second
The number of photons emitted per second (N) can be calculated using the formula:
\[ N = \frac{P}{E} \]
where \( P = 62 \, \text{W} \) (or \( 62 \, \text{J/s} \)):
\[ N = \frac{62 \, \text{J/s}}{4.14 \times 10^{-19} \, \text{J}} \]

Calculating N:
\[ N \approx 1.5 \times 10^{20} \, \text{photons/s} \]

### Final Answer
The number of photons emitted per second by the 62 W source of monochromatic light of wavelength 4800 Å is approximately:
\[ N \approx 1.5 \times 10^{20} \, \text{photons/s} \]

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