A bulb is emitted electromagnetic radiation of 660 nm wave length.
The Total energy of radiation is 3 x `10^(-18) J `The number of
emitted photon will be : `(h=6.6xx10^(-34)J xx s, C=3xx10^(8)m//s)`

Options

A. 1
B. 10
C. 100
D. 1000

Explanation

To solve the problem of finding the number of emitted photons from a bulb that emits electromagnetic radiation of wavelength 660 nm with a total energy of 3 x 10^(-18) J, we will follow these steps:

### Step 1: Convert Wavelength to Meters
The wavelength (λ) is given in nanometers (nm). We need to convert it to meters (m) for our calculations.
\[ \lambda = 660 \, \text{nm} = 660 \times 10^{-9} \, \text{m} \]

### Step 2: Use the Planck-Einstein Relation
The energy (E) of a single photon can be expressed using the Planck-Einstein relation:
\[ E = n \cdot h \cdot \nu \]
where:
- \(E\) is the total energy,
- \(n\) is the number of photons,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{J s}\)),
- \(\nu\) is the frequency.

### Step 3: Calculate Frequency
We can relate frequency (\(\nu\)) to wavelength (\(\lambda\)) using the equation:
\[ \nu = \frac{c}{\lambda} \]
where \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)). Now substituting the value of \(\lambda\):
\[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{660 \times 10^{-9} \, \text{m}} = \frac{3 \times 10^8}{660 \times 10^{-9}} \approx 4.55 \times 10^{14} \, \text{Hz} \]

### Step 4: Substitute Values into the Energy Equation
Now we can substitute the values into the energy equation:
\[ E = n \cdot h \cdot \nu \]
Rearranging gives:
\[ n = \frac{E}{h \cdot \nu} \]
Substituting the known values:
- \(E = 3 \times 10^{-18} \, \text{J}\)
- \(h = 6.63 \times 10^{-34} \, \text{J s}\)
- \(\nu \approx 4.55 \times 10^{14} \, \text{Hz}\)

\[ n = \frac{3 \times 10^{-18}}{6.63 \times 10^{-34} \cdot 4.55 \times 10^{14}} \]

### Step 5: Calculate the Number of Photons
Calculating the denominator:
\[ h \cdot \nu = 6.63 \times 10^{-34} \cdot 4.55 \times 10^{14} \approx 3.02 \times 10^{-19} \, \text{J} \]
Now substituting back to find \(n\):
\[ n = \frac{3 \times 10^{-18}}{3.02 \times 10^{-19}} \approx 9.93 \approx 10 \]

### Conclusion
Thus, the number of emitted photons is approximately **10**.

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