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

$$E = hu = \frac{hc}{\lambda}$$

where

• $E$ – photon energy (J)
• $h$ – Planck’s constant $6.626\times10^{-34}\ \text{J·s}$
• $u$ – 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 $E = hu$ 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:

$$K_{\text{max}} = hu - \phi$$

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 $u_0$:

$$\phi = hu_0$$

If the incident light has $u < u_0$, no electrons are emitted regardless of intensity.

5. Summary Table

Quantity	Symbol	Expression	Units
Photon Energy	$E$	$hu = \dfrac{hc}{\lambda}$	J (or eV)
Photon Momentum	$p$	$\dfrac{h}{\lambda} = \dfrac{E}{c}$	kg·m·s⁻¹
Work Function	$\phi$	$hu_0$	J (or eV)
Maximum Kinetic Energy of Photoelectron	$K_{\text{max}}$	$hu - \phi$	J (or eV)

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 $K_{\text{max}}$.