A certain transition emits `6.37 xx 10^(15)` quanta per second per square meter. Calculate the power output in joule square metre per second. Given `lambda = 632.8 nm`

Options

A. `5xx10^(-5)j m^(-2)s^(-1)`
B. `2xx10^(-3)j m^(-2) s^(-1)`
C. `9.2 xx10^(-3) jm^(-2)s^(-1)`
D. `4.6 xx10^(-6) j m^(-2) s^(-1)`

Answer

To calculate the power output in joules per square meter per second for the given transition that emits \(6.37 \times 10^{15}\) quanta per second per square meter, we can follow these steps:

### Step 1: Understand the relationship between power, energy, and quanta
Power (\(P\)) is defined as the energy emitted per unit area per unit time. The energy of one quantum (photon) is given by the formula:
\[ E = h \nu \]
where \(h\) is Planck's constant and \(\nu\) (nu) is the frequency of the radiation.

### Step 2: Calculate the frequency (\(\nu\))
The frequency can be calculated using the speed of light (\(c\)) and the wavelength (\(\lambda\)):
\[ \nu = \frac{c}{\lambda} \]
Given:
- \(c = 3 \times 10^8 \, \text{m/s}\)
- \(\lambda = 632.8 \, \text{nm} = 632.8 \times 10^{-9} \, \text{m}\)

Now substituting the values:
\[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{632.8 \times 10^{-9} \, \text{m}} \]

### Step 3: Perform the calculation for frequency
Calculating \(\nu\):
\[ \nu = \frac{3 \times 10^8}{632.8 \times 10^{-9}} \approx 4.74 \times 10^{14} \, \text{Hz} \]

### Step 4: Calculate the energy of one quantum
Using Planck's constant (\(h = 6.626 \times 10^{-34} \, \text{J s}\)):
\[ E = h \nu = (6.626 \times 10^{-34} \, \text{J s}) \times (4.74 \times 10^{14} \, \text{Hz}) \]

### Step 5: Perform the calculation for energy
Calculating \(E\):
\[ E \approx 3.14 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate the total energy emitted per second
The total energy emitted per second for \(n\) quanta is:
\[ \text{Total Energy} = n \cdot E \]
Given \(n = 6.37 \times 10^{15}\):
\[ \text{Total Energy} = (6.37 \times 10^{15}) \cdot (3.14 \times 10^{-19} \, \text{J}) \]

### Step 7: Perform the calculation for total energy
Calculating the total energy:
\[ \text{Total Energy} \approx 2.00 \times 10^{-3} \, \text{J} \]

### Step 8: Conclusion
Thus, the power output in joules per square meter per second is:
\[ P \approx 2.00 \times 10^{-3} \, \text{J/m}^2/\text{s} \]