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! 🎉