Question

Monochromatic light of frequency $6\times { 10 }^{ 14 }Hz$ is produced by a laser. The power emitted is $2\times { 10 }^{ -3 }W$.
A. What is the energy of the photon in the light beam?
B. How many photons per second, on an average are emitted by the source?

Answer 1

To calculate the energy of the photon, use energy-frequency relation. The ratio of the power emitted and the energy of the photon gives the number of photons emitted per second by the source.

Complete answer:
Given: Frequency of light ($\nu$)= $6\times { 10 }^{ 14 }Hz$
Power emitted (P) = $2\times { 10 }^{ -3 }W$

A. Energy for a photon is given by,
$E\quad =\quad h\nu$ …(1)
where, h: Planck’s constant which is equal to $6.63\times { 10 }^{ -34 }\quad J.s$
By substituting values in the equation. (1) we get,
$E\quad = 6.63\times { 10 }^{ -34 }\times 6\times { 10 }^{ 14 }$
$\Rightarrow E=39.78\times { 10 }^{ -20 }$
Thus, the energy of the photon in a light beam is $39.78\times { 10 }^{ -20 }$.

B. Photons emitted per second is given by,
$n = \dfrac { Power }{ Energy }$
where, n: photons emitted for second
Substituting the values in above equation,
$n= \dfrac { 2\quad \times { 10 }^{ -3 } }{ 39.78 \times { 10 }^{ -20 } }$
$\Rightarrow n = 0.05 \times { 10 }^{ 17 }$
$\Rightarrow n = 5 \times { 10 }^{ 15 }$
Thus, on an average $5 \times { 10 }^{ 15 }$ photons per second are emitted by the source.

Note:
If instead of frequency, the wavelength was mentioned in the question, still the problem would be easy to solve. In that case, the formula used will be,
$E = \dfrac { hc }{\lambda }$
Substituting the values in the above equation will give the energy of the photon in the light beam.