MP Board · Class 12 · Physics · Dual Nature of Radiation and MatterState Einstein's photoelectric equation and explain how it successfully explains the laws of photoelectric emission. Also, solve the following numerical problem: Ultraviolet light of wavelength 2271 Å is irradiated on a work function of a photocathode. If the stopping potential is 1.3 V, calculate the work function of the metal in electron volts (eV).

Einstein's Photoelectric Equation and Explanation

\nAlbert Einstein explained the photoelectric effect in 1905 using Max Planck's quantum theory. According to Einstein, light consists of energy packets called photons, each having an energy of $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of light.

Key Principles:

• Energy Transfer: When a photon of energy $h\nu$ collides with an electron on the metal surface, it transfers its entire energy to the electron in a single instantaneous interaction.
• Work Function: A minimum amount of energy, called the work function ($\phi_0$ or $W_0$), is required by the electron to escape the metal surface.
• Maximum Kinetic Energy: The remaining energy of the photon appears as the maximum kinetic energy ($K_{max}$) of the emitted photoelectron. \nThus, Einstein's photoelectric equation is given by: $$K_{max} = h\nu - \phi_0$$\nOR in terms of stopping potential ($V_0$): $$e V_0 = h\nu - \phi_0$$

Explanation of Laws of Photoelectric Emission:

1. Effect of Frequency: If the photon energy $h\nu$ is less than the work function $\phi_0$, emission is impossible. The minimum frequency required is threshold frequency ($\nu_0$). Higher frequency increases $K_{max}$.
2. Effect of Intensity: Increasing the intensity of light increases the number of incident photons, which in turn increases the number of emitted photoelectrons (current), but does not affect their kinetic energy.
3. Instantaneous Process: The collision between a photon and an electron is instantaneous, meaning there is no time lag between the incidence of light and the emission of photoelectrons.

Numerical Solution

Given Data:

• Wavelength of incident light, $\lambda = 2271 \text{ Å} = 2271 \times 10^{-10} \text{ m}$
• Stopping potential, $V_0 = 1.3 \text{ V}$
• Planck's constant, $h \approx 6.63 \times 10^{-34} \text{ J s}$
• Speed of light, $c = 3 \times 10^8 \text{ m/s}$
• Charge of electron, $e = 1.6 \times 10^{-19} \text{ C}$

Step 1: Calculate the energy of the incident photon ($E$)

$$E = \frac{hc}{\lambda}$$ $$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2271 \times 10^{-10}}$$ $$E = \frac{1.989 \times 10^{-25}}{2271 \times 10^{-10}} \approx 8.7586 \times 10^{-19} \text{ Joules}$$ \nConverting energy into electron volts (eV): $$E_{\text{eV}} = \frac{8.7586 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 5.474 \text{ eV}$$

Step 2: Calculate maximum kinetic energy ($K_{max}$)

$$K_{max} = e V_0$$ $$K_{max} = 1.3 \text{ eV} \quad (\text{since } 1 \text{ V} \times e = 1 \text{ eV})$$

Step 3: Calculate the work function ($\phi_0$)\nUsing Einstein's equation:

$$K_{max} = E - \phi_0$$ $$\phi_0 = E - K_{max}$$ $$\phi_0 = 5.474 \text{ eV} - 1.3 \text{ eV}$$ $$\phi_0 = 4.174 \text{ eV}$$

Answer:\nThe work function of the metal is approximately 4.17 eV.