explain photoelectric emission in terms of photon energy and work function energy

Energy and Momentum of a Photon

1. Photon Energy

The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. It is given by

\$E = h\nu = \frac{hc}{\lambda}\$

where

• \$E\$ – photon energy (J)
• \$h\$ – Planck’s constant \$6.626\times10^{-34}\ \text{J·s}\$
• \$\nu\$ – frequency (Hz)
• \$c\$ – speed of light \$3.00\times10^{8}\ \text{m·s}^{-1}\$
• \$\lambda\$ – wavelength (m)

2. Photon Momentum

Even though a photon has no rest mass, it carries momentum. The magnitude of the momentum is

\$p = \frac{E}{c} = \frac{h}{\lambda}\$

with \$p\$ measured in kg·m·s⁻¹ (or N·s). This relationship will be useful when discussing conservation of momentum in the photoelectric effect.

3. Photoelectric Emission

When light of sufficient frequency shines on a metal surface, electrons can be ejected. The process is explained by the photon model as follows:

1. A photon of energy \$E = h\nu\$ is absorbed by an electron in the metal.
2. If the photon energy exceeds the work function \$\phi\$ of the metal, the electron can escape.
3. The kinetic energy \$K_{\text{max}}\$ of the emitted electron is the excess energy:

\$K_{\text{max}} = h\nu - \phi\$

Here \$\phi\$ is the minimum energy required to remove an electron from the surface (the work function), typically expressed in electronvolts (eV).

4. Work Function (\$\phi\$)

The work function depends on the material and its surface condition. It can be related to a threshold frequency \$\nu_0\$:

\$\phi = h\nu_0\$

If the incident light has \$\nu < \nu_0\$, no electrons are emitted regardless of intensity.

5. Summary Table

6. Practical Implications

Understanding the relationship between photon energy and work function allows us to predict:

• The minimum frequency (or maximum wavelength) of light that will cause emission for a given metal.
• The kinetic energy distribution of emitted electrons, which is measured in photoelectron spectroscopy.
• The role of intensity: increasing intensity raises the number of photons (and thus the number of emitted electrons) but does not affect \$K_{\text{max}}\$.