The number of photons emitted per second by a 60 W source of monochromatic light of wavelength 663 nm is:
`(h=6.63xx10^(-34)Js )`

Options

A. `4xx10^(-20)`
B. `1.5xx10^(20)`
C. `2xx10^(-20)`
D. `2xx10^(20)`

Explanation

To find the number of photons emitted per second by a 60 W source of monochromatic light with a wavelength of 663 nm, we can follow these steps:

### Step 1: Understand the relationship between energy, frequency, and the number of photons
The energy of one photon is given by the formula:
\[ E = h \nu \]
where:
- \( E \) is the energy of one photon,
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( \nu \) is the frequency of the light.

For \( n \) photons, the total energy emitted per second (power) is:
\[ E = n h \nu \]

### Step 2: Relate frequency and wavelength
We know that the frequency \( \nu \) can be related to the wavelength \( \lambda \) using the speed of light \( c \):
\[ c = \nu \lambda \]
Thus, we can express frequency as:
\[ \nu = \frac{c}{\lambda} \]
where:
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength (given as 663 nm).

### Step 3: Convert the wavelength to meters
Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \):
\[ \lambda = 663 \, \text{nm} = 663 \times 10^{-9} \, \text{m} \]

### Step 4: Substitute frequency into the energy equation
Substituting \( \nu \) into the energy equation gives:
\[ E = n h \left(\frac{c}{\lambda}\right) \]
Rearranging for \( n \):
\[ n = \frac{E \lambda}{h c} \]

### Step 5: Substitute known values into the equation
Now we can substitute the known values:
- \( E = 60 \, \text{W} \) (which is \( 60 \, \text{J/s} \)),
- \( \lambda = 663 \times 10^{-9} \, \text{m} \),
- \( h = 6.63 \times 10^{-34} \, \text{Js} \),
- \( c = 3 \times 10^8 \, \text{m/s} \).

So we have:
\[ n = \frac{60 \times (663 \times 10^{-9})}{(6.63 \times 10^{-34}) \times (3 \times 10^8)} \]

### Step 6: Calculate the number of photons
Calculating the numerator:
\[ 60 \times 663 \times 10^{-9} = 39780 \times 10^{-9} = 3.978 \times 10^{-5} \]

Calculating the denominator:
\[ (6.63 \times 10^{-34}) \times (3 \times 10^8) = 1.989 \times 10^{-25} \]

Now substituting these values into the equation for \( n \):
\[ n = \frac{3.978 \times 10^{-5}}{1.989 \times 10^{-25}} \]

Calculating \( n \):
\[ n \approx 2.00 \times 10^{20} \]

### Final Answer
The number of photons emitted per second by the 60 W source of monochromatic light of wavelength 663 nm is approximately:
\[ n \approx 2 \times 10^{20} \]

---