Question

Intensity of a LASER beam$\left( {\lambda = 6600\mathop A\limits^0 } \right)$ is \[6mw\]. Calculate the number of photons passing along the path of the beam in per second.

Answer 1

Hint:Einstein explained the photoelectric effect on the basis of Planck’s quantum theory of radiation. According to him, photoelectric emission does not take place by continuous absorption of energy from radiation. Light radiation consists of discrete packets of energy called photons having energy$E = h\nu $

Complete step by step answer:
Given: Wavelength of the laser beam,
\[ \Rightarrow \lambda = 6600angstorm\]
$ \Rightarrow 6600 \times {10^{ - 10}}m$
Energy of each photon in the light beam is given by,
$E = h\nu $……………………(1)
Where, $h$ is the Planck’s constant
\[\upsilon \]is the frequency of a laser beam.
We know that, $c = \nu \lambda $
$ \Rightarrow \nu = \dfrac{c}{\lambda }$
Here $C$ is the speed of light.
Equation (1) can also be written as,
$ \Rightarrow E = h\dfrac{c}{\lambda }$
Now substitute the given values, we get
$ \Rightarrow E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6600 \times {{10}^{ - 10}}}}$
$ \Rightarrow E = 3 \times {10^{ - 19}}J$
Power or intensity of a laser beam,
$ \Rightarrow P = 6mW$
$ \Rightarrow 6 \times {10^{ - 3}}W$
Now, the number of photons falling per second on the target is given by,
$ \Rightarrow n = \dfrac{P}{E}$
Substitute the values we get
$ \Rightarrow n = \dfrac{{6 \times {{10}^{ - 3}}W}}{{3 \times {{10}^{ - 19}}J}}$
$\therefore n = 2 \times {10^{16}}$ Photons per second.
Hence, the number of photons passing along path of beam in per second is $2 \times {10^{16}}$

Additional information:
According to Einstein’s photoelectric equation,
When a light is incident on a metal, the photons having energy \[h\nu \] collide with electrons at the surface of the metal. During these collisions, the energy of the photon is completely transferred to the electron. If this energy is sufficient, the electrons are ejected out of the metal instantaneously .The minimum energy needed for the electron to come out of the metal surface is called work function. If the energy \[h\nu \] of the incident photon exceeds the work function \[\left( {{\phi _0}} \right)\] ,the electrons are emitted with a maximum kinetic energy.
\[ \Rightarrow h\nu = {f_0} + {K_{\max }}\]
\[ \Rightarrow {K_{\max }} = h\nu - {f_0}\]

Note:
-For a given metal and frequency of incident radiation the photoelectric current is directly proportional to the intensity of incident radiation.
-The photoelectric effect is an instantaneous process.
-The maximum kinetic energy of the emitted photoelectrons is independent of the intensity of the incident radiation but depends only on the frequency of the incident radiation.