Question

Monochromatic light of wavelength 3000 ${A^0}$ is incident on a surface area 4 $c{m^2}$. If intensity of light is 150$mW/{m^2}$ , then rate at which photons strike the target is
(A). $3 \times {10^{10}}/s$
(B). $9 \times {10^{13}}/s$
(C). $7 \times {10^{15}}/s$
(D). $6 \times {10^{19}}/s$

Answer 1

- Hint: In order to find the rate of photons, we have to calculate the total energy that is incident on the surface and the energy of each photon. Every photon has the same energy, so the total energy will be distributed equally to all the photons. Thus by dividing the total energy to that of energy of one photon gives us the number of photons.

Formula used
$I = \dfrac{P}{A}$ , where $I$ denotes the intensity,$P$ denotes power and $A$ denotes the area measured on a plane which is perpendicular to the direction of flow of energy.

${E_P} = \dfrac{{h \times c}}{\lambda }$ , where ${E_p}$the energy of a single photon is,$\lambda$ is the wavelength of light in meters ,$c$is the speed of light in free space and $h$ is the Planck’s constant.

Complete step-by-step solution -
Intensity of radiant energy is defined as the total power transmitted per unit area, where area is measured in a plane which is perpendicular to the direction of propagation of energy, i.e. $I = \dfrac{P}{A}$.
Substituting $I = 150 \times {10^{ - 3}}W/{m^2}$ and area $A = 4 \times {10^{ - 4}}{m^2}$ we can calculate power as $P = I \times A$
Therefore power $P = 150 \times {10^{ - 3}} \times 4 \times {10^{ - 4}}$
$P = 6 \times {10^{ - 5}}W$
For one second ${E_T} = 6 \times {10^{ - 5}}W$ .

We know from Planck’s energy-frequency relation that ${E_P} = h\nu$ where ${E_p}$is the energy of photon (also known as photon energy) , $I$is the Planck’s constant whose value is $6.626 \times {10^{ - 34}}J.s$ in SI units and $\nu$ is the frequency of the wave. Also note that the frequency of a wave can be expressed in terms of its wavelength and speed. This is given by the equation $\nu = \dfrac{c}{\lambda }$.Here $c$ is the speed of light ($3 \times {10^8}m/s$) and $\lambda$ is the wavelength. Substituting this equation in Planck’s energy-frequency relation we get ${E_P} = \dfrac{{h \times c}}{\lambda }$ .Given that $\lambda = 3000{A^0}$
Therefore ${E_P} = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3000 \times {{10}^{ - 10}}}}$
${E_P} = 6.63 \times {10^{ - 19}}J$
Taking the ratio of both, we get the number of photons
$n = \dfrac{{{E_T}}}{{{E_P}}}$
$n = \dfrac{{6 \times {{10}^{ - 5}}}}{{6.63 \times {{10}^{ - 19}}}}$

$n = 9 \times {10^{13}}photons/\sec ond$
The correct option is B

Note: We know power is defined as the work/energy per unit time. So for 1 second power and energy will be equal. This is an instantaneous process. In photoelectric effect the number of electrons emitted is affected by the intensity of the light falling on the surface. The frequency of light does not affect the number of electrons. And the intensity does not affect the kinetic energy of the photons.