recall and use hf = Φ + 21mvmax2

⚡ Energy and Momentum of a Photon

A photon is a packet of light energy. Its energy depends only on its frequency (\$f\$) and is given by Planck’s relation:

\$E_{photon}=hf\$

where \$h=6.63\times10^{-34}\text{ J·s}\$ is Planck’s constant.

When a photon strikes a metal surface, it can eject an electron if its energy exceeds the work function (\$\Phi\$) of the metal. The excess energy becomes the kinetic energy of the emitted electron.

📋 Photoelectric Effect Equation

The conservation of energy for the photoelectric process is written as:

\$hf = \Phi + \frac{1}{2}mv_{max}^2\$

• \$hf\$ – energy of the incoming photon
• \$\Phi\$ – work function (minimum energy needed to free an electron from the metal)
• \$\frac{1}{2}mv_{max}^2\$ – maximum kinetic energy of the ejected electron
• \$m\$ – electron mass (\$9.11\times10^{-31}\text{ kg}\$)
• \$v_{max}\$ – maximum speed of the emitted electron

🔢 Symbols Reference

🧮 Worked Example

A photon of frequency \$f = 1.0\times10^{15}\text{ Hz}\$ strikes a sodium surface whose work function is \$\Phi = 2.3\text{ eV}\$. Calculate the maximum speed of the emitted electron.

1. Convert work function to joules:

   \$\Phi = 2.3\text{ eV}\times1.60\times10^{-19}\frac{\text{J}}{\text{eV}} = 3.68\times10^{-19}\text{ J}\$

2. Photon energy:

   \$E_{photon}=hf = (6.63\times10^{-34})(1.0\times10^{15}) = 6.63\times10^{-19}\text{ J}\$

3. Kinetic energy of electron:

   \$K{max}=E{photon}-\Phi = 6.63\times10^{-19} - 3.68\times10^{-19}=2.95\times10^{-19}\text{ J}\$

4. Use \$K{max}=\frac{1}{2}mv{max}^2\$ to find \$v_{max}\$:

   \$v{max}=\sqrt{\frac{2K{max}}{m}} = \sqrt{\frac{2(2.95\times10^{-19})}{9.11\times10^{-31}}}\$

   \$v_{max}\approx \sqrt{6.48\times10^{11}} \approx 8.0\times10^{5}\text{ m/s}\$

Thus the fastest electron leaves the sodium surface with a speed of about \$8.0\times10^{5}\text{ m/s}\$.

✏️ Practice Problems

1. A photon of wavelength \$400\text{ nm}\$ hits a metal with work function \$2.0\text{ eV}\$. Find the maximum kinetic energy of the emitted electron in electron‑volts. (Hint: \$E=hc/\lambda\$, \$c=3.00\times10^{8}\text{ m/s}\$.)
2. If the stopping potential needed to halt the most energetic electrons is \$1.5\text{ V}\$, what is the work function of the metal when illuminated by light of frequency \$5.0\times10^{14}\text{ Hz}\$?
3. Explain why increasing the intensity of the light (while keeping frequency constant) does not change \$v_{max}\$ of the photoelectrons.

🎉 Keep practicing – you’ve got this! 🎉