Question:
The average number of photons emitted per second by a laser of power \(6.6\times10^{-3}\ \text{W}\) producing a light of wavelength \(600\ \text{nm}\) is
\[
\text{(Planck's constant, } h=6.6\times10^{-34}\ \text{J s)}
\]
The average number of photons emitted per second by a laser of power \(6.6\times10^{-3}\ \text{W}\) producing a light of wavelength \(600\ \text{nm}\) is
\[ \text{(Planck's constant, } h=6.6\times10^{-34}\ \text{J s)} \]
\[ \text{(Planck's constant, } h=6.6\times10^{-34}\ \text{J s)} \]
Show Hint
For photon emission problems, first calculate the energy of a single photon using
\[
E=\frac{hc}{\lambda}.
\]
Then use
\[
P=NE
\]
to determine the number of photons emitted per second.
Updated On: Jun 26, 2026
- \(2\times10^{16}\)
- \(3\times10^{16}\)
- \(4\times10^{16}\)
- \(6\times10^{16}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the relation between power and number of photons emitted.
Power is the energy emitted per unit time.
If \(N\) photons are emitted per second and each photon has energy \(E\), then \[ P=NE \] Hence, \[ N=\frac{P}{E} \]
Step 2: Calculate the energy of one photon.
The energy of a photon is given by \[ E=\frac{hc}{\lambda} \] Given, \[ h=6.6\times10^{-34}\ \text{J s} \] \[ c=3\times10^{8}\ \text{m s}^{-1} \] \[ \lambda=600\ \text{nm} =600\times10^{-9}\ \text{m} \] Substituting, \[ E= \frac{(6.6\times10^{-34})(3\times10^{8})} {600\times10^{-9}} \] \[ E= \frac{19.8\times10^{-26}} {6\times10^{-7}} \] \[ E=3.3\times10^{-19}\ \text{J} \]
Step 3: Calculate the number of photons emitted per second.
Given laser power, \[ P=6.6\times10^{-3}\ \text{W} \] Therefore, \[ N=\frac{6.6\times10^{-3}} {3.3\times10^{-19}} \] \[ N=2\times10^{16} \]
Step 4: Final conclusion.
Hence, the average number of photons emitted per second is \[ \boxed{2\times10^{16}} \]
Power is the energy emitted per unit time.
If \(N\) photons are emitted per second and each photon has energy \(E\), then \[ P=NE \] Hence, \[ N=\frac{P}{E} \]
Step 2: Calculate the energy of one photon.
The energy of a photon is given by \[ E=\frac{hc}{\lambda} \] Given, \[ h=6.6\times10^{-34}\ \text{J s} \] \[ c=3\times10^{8}\ \text{m s}^{-1} \] \[ \lambda=600\ \text{nm} =600\times10^{-9}\ \text{m} \] Substituting, \[ E= \frac{(6.6\times10^{-34})(3\times10^{8})} {600\times10^{-9}} \] \[ E= \frac{19.8\times10^{-26}} {6\times10^{-7}} \] \[ E=3.3\times10^{-19}\ \text{J} \]
Step 3: Calculate the number of photons emitted per second.
Given laser power, \[ P=6.6\times10^{-3}\ \text{W} \] Therefore, \[ N=\frac{6.6\times10^{-3}} {3.3\times10^{-19}} \] \[ N=2\times10^{16} \]
Step 4: Final conclusion.
Hence, the average number of photons emitted per second is \[ \boxed{2\times10^{16}} \]
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