To find out how many photons are emitted per second by the laser, we need to calculate the energy of a single photon and then determine how many such photons correspond to the total power emitted.
The energy \(E\) of a single photon can be determined using Planck's equation:
\(E = h \times f\)
Where:
Substituting the given values into the formula:
\(E = 6.63 \times 10^{-34} \times 6 \times 10^{14}\)
\(E = 3.978 \times 10^{-19} \, \text{J}\)
This is the energy of one photon. The total power \(P\) emitted by the laser is \(2 \times 10^{-3} \, \text{W}\) (which is joules per second).
We can find the number of photons emitted per second by dividing the total power by the energy of a single photon:
\(n = \frac{P}{E} = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}}\)
Calculating this, we find:
\(n \approx 5.03 \times 10^{15}\)
Therefore, the number of photons emitted per second is approximately \(5 \times 10^{15}\).
Thus, the correct answer is \(5 \times 10^{15}\), which matches with one of the given options.
Given: - Frequency of light: \( \nu = 6 \times 10^{14} \, \text{Hz} \) - Power emitted by the source: \( P = 2 \times 10^{-3} \, \text{W} \) - Planck's constant: \( h = 6.63 \times 10^{-34} \, \text{Js} \)
The energy \( E \) of a photon is given by:
\[ E = h\nu \]
Substituting the given values:
\[ E = 6.63 \times 10^{-34} \times 6 \times 10^{14} \, \text{J} \] \[ E = 3.978 \times 10^{-19} \, \text{J} \]
Rounding off:
\[ E \approx 4 \times 10^{-19} \, \text{J} \]
The number of photons emitted per second (\( n \)) is given by:
\[ n = \frac{P}{E} \]
Substituting the given values:
\[ n = \frac{2 \times 10^{-3}}{4 \times 10^{-19}} \] \[ n = \frac{2}{4} \times 10^{16} \] \[ n = 0.5 \times 10^{16} \] \[ n = 5 \times 10^{15} \]
Conclusion: The number of photons emitted per second by the source is \( 5 \times 10^{15} \).
Match the LIST-I with LIST-II:
| List-I | List-II | ||
| A. | Radio-wave | I. | is produced by Magnetron valve |
| B. | Micro-wave | II. | due to change in the vibrational modes of atoms |
| C. | Infrared-wave | III. | due to inner shell electrons moving from higher energy level to lower energy level |
| D. | X-ray | IV. | due to rapid acceleration of electrons |
Choose the correct answer from the options given below:

In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 