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Photoelectric Effect

CSCA Photoelectric Effect study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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Tutorial Content

Modern Physics: Photoelectric Effect

1. Phenomenon & Definition

The **Photoelectric Effect** is the emission of electrons when light shines on a material. The emitted electrons are called **photoelectrons**, and the resulting current is the **photocurrent**.

Classical Dilemma vs. Quantum Explanation

* **Classical Prediction (Fail)**: Energy depends on **intensity**. Strong enough light should eject electrons at any frequency. This contradicted experiments.

* **Einstein's Photon Theory (Success)**:

1. Light consists of particles (photons), each with energy $E = h\nu$.

2. The effect is an **instantaneous, one-on-1 collision** between a photon and an electron.

3. Electrons must overcome the metal's binding energy, called the **Work Function ($W_0$)**.

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2. Core Equation: Einstein's Photoelectric Equation

This is the most critical formula for calculation problems in the CSCA exam:

$$ E_{k\text{max}} = h\nu - W_0 $$

Or expressed with wavelength:

$$ \frac{1}{2}mv^2_{\text{max}} = h\frac{c}{\lambda} - W_0 $$

**Variables Definitions**:

* $h\nu$: **Photon Energy**. Determined by frequency $\nu$ or wavelength $\lambda$.

* $W_0$: **Work Function**. An intrinsic property of the metal. If $h\nu < W_0$, no effect occurs.

* $E_{k\text{max}}$: **Maximum Kinetic Energy** of the photoelectron.

**Key Deduction**:

* **Cutoff Frequency ($\nu_0$)**: The minimum frequency required to eject an electron. Defined when $E_{k\text{max}}=0$, so $h\nu_0 = W_0$.

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3. Graphical Analysis (Exam Focus)

(1) Max Kinetic Energy vs. Frequency ($E_{k\text{max}} - \nu$)

This follows a linear equation: $y = kx + b \rightarrow E_{k\text{max}} = h\nu - W_0$.

* **Slope**: Equals **Planck's constant $h$**. Lines for different metals are parallel.

* **X-intercept**: **Cutoff Frequency $\nu_0$**.

* **Y-intercept**: Negative Work Function $-W_0$.

(2) Current vs. Voltage ($I - U$)

Shows how photocurrent changes with applied voltage. Two key points:

* **Saturation Current ($I_m$)**: The maximum current when voltage is high enough to collect all electrons. $I_m$ is proportional to **Light Intensity**.

* **Stopping Potential ($U_c$)**: The reverse voltage needed to stop the current. Work done by the field equals max kinetic energy:

$$ eU_c = E_{k\text{max}} $$

**Critical Point**: $U_c$ depends **ONLY on light frequency**, not intensity! Higher frequency = Higher stopping potential.

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4. CSCA Strategy & Tips

1. **Calculation Shortcuts**:

* Planck's constant is usually given as $h = 6.63 \times 10^{-34} \text{J}\cdot\text{s}$.

* However, for electron-volts (eV), memorize this combination:

$$ hc \approx 1240 \, \text{eV}\cdot\text{nm} $$

* **Example**: For $\lambda = 400\text{nm}$, Photon Energy $E = \frac{1240}{400} = 3.1 \, \text{eV}$. Much faster than using Joules.

2. **Common Traps**:

* **Intensity Increases** $\rightarrow$ More photons $\rightarrow$ More electrons $\rightarrow$ Higher Saturation Current ($I_m$).

* **Intensity Increases** $\rightarrow$ Photon energy **SAME** $\rightarrow$ Max Kinetic Energy ($E_{k\text{max}}$) **SAME** $\rightarrow$ Stopping Potential ($U_c$) **SAME**.

5. Typical Example

**Problem**: Light of frequency $\nu$ hits a metal with work function $W_0$, yielding stopping potential $U_c$. If the frequency is doubled ($2\nu$), what is the new stopping potential?

**Solution**:

1. Initial: $eU_c = h\nu - W_0$

2. Final: $eU'_c = h(2\nu) - W_0$

3. Compare: $eU'_c = 2h\nu - W_0 = 2(eU_c + W_0) - W_0 = 2eU_c + W_0$

4. Conclusion: $U'_c = 2U_c + \frac{W_0}{e}$. The new stopping potential is **more than** double the original.