Question

Calculate number of photons passing through a ring of unit area in unit time. If light of intensity $100\,W{m^2}$ and of wavelength $400\,nm$ is falling normally on the ring.

Answer 1

Use the Einstein’s formula to find the energy of the photon by substituting the known values and constants in it. With the help of the obtained energy value and the intensity, use the formula of the number of photons to find the number.

Formula used:
(1) The energy of the photon is given by
$E = \dfrac{{hc}}{\lambda }$
Where $E$ is the energy of the photon, $h$ is the Planck’s constant, $c$ is the velocity of the light and $\lambda$ is the wavelength of the light.
(2) Number of photons is given by
$N = \dfrac{I}{E}$
Where $N$ is the number of photons passing through the particular ring, $I$ is the intensity of the light passing through the ring.

Complete step by step answer:
It is given that the intensity of the light, $I = 100\,W{m^2}$
Wavelength of the light, $\lambda = 400\,nm = 400 \times {10^{ - 9}}\,m$
Using the formula of the energy of the photon,
$E = \dfrac{{hc}}{\lambda }$
Substituting the known values in the above equation and also taking the Planck’s constant as $6.6 \times {10^{ - 34}}$, we get
$E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{400 \times {{10}^{ - 9}}}}$
By performing the basic arithmetic operation, we get
$E = 4.95 \times {10^{ - 19}}\,J$
Hence the energy of the photons in the light is obtained as $4.95 \times {10^{ - 19}}\,J$ .
Use the formula of the number of photons,
$N = \dfrac{I}{E}$
Substitute the obtained energy of the photon and the intensity of the light which passes the ring,
$N = \dfrac{{100}}{{4.95 \times {{10}^{ - 19}}}}$
By simplifying the above step, we get
$N = 2 \times {10^{20}}$ photons per second.

Note: The photons mentioned in this solution, is the tiny particle that has no mass and also no charge but it is considered as the type of the electric field that passes through space. They have the ability to travel at the light speed.