Question

State Einstein's photoelectric equation and mention physical significance of each term involved in it. The wavelength of incident light is 4000Å. Calculate the energy of incident photon.

Hint:

Think of it as a transaction: The photon gives its energy (\(h\nu\)) to the electron. The electron pays a "tax" or "exit fee" (\(\phi_0\)) to leave the metal. The leftover money is its kinetic energy (\(K_{max}\)).

Answer 1

Einstein's Photoelectric Equation: The equation is: \[ K_{max} = h\nu - \phi_0 \] or \[ h\nu = \phi_0 + K_{max} \] Physical Significance of Each Term: \(h\nu\): This term represents the energy of the incident photon. According to Planck's quantum theory, light consists of discrete packets of energy called photons. The energy of each photon is proportional to its frequency (\(\nu\)), where \(h\) is Planck's constant. This is the total energy supplied to a single electron on the metal surface.
\(\phi_0\): This is the work function of the metal. It represents the minimum amount of energy required to just liberate an electron from the surface of the metal, overcoming the attractive forces holding it. The work function is a characteristic property of the material.
\(K_{max}\): This is the maximum kinetic energy of the emitted electron (photoelectron). It represents the excess energy from the photon that is converted into the electron's kinetic energy after it has escaped the surface. The kinetic energy is a maximum because some electrons deeper within the metal may lose energy through collisions before they escape.
Calculation:
We need to find the energy (\(E\)) of the incident photon. Given:
Wavelength of light, \(\lambda = 4000 \, \AA = 4000 \times 10^{-10} \, m = 4 \times 10^{-7} \, m\)
Constants:
Planck's constant, \(h = 6.63 \times 10^{-34} \, J \cdot s\)
Speed of light, \(c = 3 \times 10^8 \, m/s\)
The energy of a photon is given by the formula: \[ E = h\nu = \frac{hc}{\lambda} \] \[ E = \frac{(6.63 \times 10^{-34} \, J \cdot s)(3 \times 10^8 \, m/s)}{4 \times 10^{-7} \, m} \] \[ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} \, J \] \[ E = 4.9725 \times 10^{-19} \, J \] The energy of the incident photon is \(4.97 \times 10^{-19}\) Joules.