Derive Einstein's Photoelectric Equation and Verify Laws of Photoelectric emission using this equation.

Derive Einstein's Photoelectric Equation and Verify Laws of Photoelectric emission using this equation.

According to Einstein, when a photon of energy $h\nu$ falls on a metal surface, the energy of the photon is used in two ways:

(i) A part of photon energy is used by an electron to just cross over the surface barrier so that it may come out of the metal surface. This part of energy is equal to the work function ($\phi_0$) of the metal ;

(ii) The remaining part of the photon energy is used in giving a velocity to the emitted photo electron. This part of energy is equal to the kinetic energy ($\frac{1}{2}mv^2$) of the emitted photo-electron. The process is shown in figure 

According to the law of conservation of energy,

$$h\nu = \phi_0 + \frac{1}{2}mv^2 \quad \dots (i)$$

If $\nu$ is just equal to threshold frequency $\nu_0$, then the free electron will just come out of the surface of metal and its kinetic energy is zero.

i.e.,

$$h\nu_0 = \phi_0 + 0$$

$$h\nu_0 = \phi_0 \quad \dots (ii)$$

Hence, eqn. $(i)$ can be written as,

$$h\nu = h\nu_0 + \frac{1}{2}mv^2$$

or

$$\frac{1}{2}mv^2 = h\nu - h\nu_0$$

$$\frac{1}{2}mv^2 = h(\nu - \nu_0) \quad \dots (iii)$$

which is the expression for the work function of the metal.

Work function, $\phi_0 = h\nu_0 = \frac{hc}{\lambda_0}$, where $\lambda_0$ is called threshold wavelength for the photo sensitive material.

Substituting the value of eqn. $(ii)$ in eqn. $(i)$, we get,

$$h\nu = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2$$

or

$$\frac{1}{2}mv^2 = h\nu - \frac{hc}{\lambda_0}$$

Equation $(iii)$ is called Einstein's photoelectric equation. 

This equation shows that the kinetic energy ($\frac{1}{2}mv^2$) of a photo electron is directly proportional to the frequency ($\nu$) of the incident radiation. In other words, kinetic energy of a photoelectron is inversely proportional to the wavelength of the incident radiation.

Graph Between K.E of Emitted Photoelectrons and frequency of incident radiation is 

Verifications of the Laws of Photoelectric Emission using Einstein's Photoelectric Equation : 

(i) If incident frequency $\nu <$ threshold frequency ($\nu_0$), then kinetic energy ($\frac{1}{2}mv^2$) of photo electron $= h(\nu - \nu_0)$ is negative. This is not possible because kinetic energy cannot be negative. Hence, photoelectric emission is not possible if frequency ($\nu$) of incident light is less than the threshold frequency ($\nu_0$) of the metal.

(ii) One photon can emit only one electron from the metal surface, so the number of photo-electrons emitted per second is directly proportional to the intensity of incident light which depends upon number of photons present in the incident light, provided $\nu > \nu_0$.

(iii) Kinetic energy, $\frac{1}{2}mv^2 = h(\nu - \nu_0)$ increases with the increase in the frequency ($\nu$) of the incident light as $h$ and $\nu_0$ are constants.

(iv) Photoelectric emission is due to elastic collision between a photon and an electron in the substance as such , there cannot be any significant time lag between incidenceof photon and emission of photoelectron. Thus , the process of photo electric emission is instantaneous.