E2-2. The Photoelectric Equation and Stopping PotentialHL Only

1. The Photoelectric Equation

Einstein’s photoelectric equation relates the energy of the incident photon to the work function of the metal and the kinetic energy of the emitted photoelectron:

$$E = hf = \Phi + E_{k \text{max}}$$
  • $E$: Total energy of the incident photon (J or eV).
  • $h$: Planck's constant ($6.63 \times 10^{-34} \text{ J s}$).
  • $f$: Frequency of the incident light (Hz).
  • $\Phi$: Work function of the metal (J or eV).
  • $E_{k \text{max}}$: Maximum kinetic energy of the emitted photoelectron (J or eV).

Graphical Representation:

The equation can be rearranged into the standard straight-line form $y = mx + c$: $$E_{k \text{max}} = hf - \Phi$$ Plotting $E_{k \text{max}}$ on the y-axis against $f$ on the x-axis yields a straight line where:

  • The gradient (slope) is always equal to Planck's constant, $h$.
  • The x-intercept is the threshold frequency, $f_0$.
  • The y-intercept represents the negative work function, $-\Phi$.
Maximum Kinetic Energy vs. Frequency Frequency (f) Ek,max f₀ (Threshold) -Φ 0 Gradient = h

Exam Tip — Units in the Photoelectric Equation

When using the photoelectric equation, $hf$, $\Phi$, and $E_{k \text{max}}$ must all have the exact same units. Make sure your quantities are all unified in either electronvolts (eV) or Joules (J). To convert from Joules to eV, divide by $1.60 \times 10^{-19} \text{ J}$.

2. Stopping Potential ($V_s$)

The stopping potential ($V_s$) provides an experimental way to measure the maximum kinetic energy of emitted photoelectrons. If photoelectrons are emitted from a metal plate and directed towards another plate across a vacuum, a reverse potential difference can be applied to oppose their motion.

Experimental Setup for Stopping Potential Evacuated Quartz Tube Cathode Anode Light (hf) e⁻ e⁻ Repulsive E-Field V A + - Variable Power Supply
  • As the reverse potential increases, fewer electrons have enough kinetic energy to cross the gap.
  • Eventually, a specific potential difference is reached where zero electrons can cross the gap, bringing the photoelectric current to zero. This value is the stopping potential, $V_s$.
  • At this point, the electrical work done to stop the fastest electron is exactly equal to its maximum kinetic energy: $$E_{k \text{max}} = eV_s$$

Effects of Intensity and Frequency:

  • Changing Intensity: Increasing the intensity (brightness) of the light increases the number of incident photons per second. This increases the photoelectric current (more electrons emitted) but has no effect on the stopping potential or maximum kinetic energy, because the energy of individual photons remains unchanged.
  • Changing Frequency: Increasing the frequency of the incident light increases the energy of the individual photons ($E=hf$). This directly increases the maximum kinetic energy of the emitted electrons, thereby increasing the stopping potential required to halt them. It does not change the photoelectric current if the number of photons arriving per second is kept constant.
Photocurrent vs. Voltage Characteristics Potential Difference (V) Photocurrent (I) 0 Reverse Bias Forward Bias -V_{s1} Higher Intensity (Same f) Lower Intensity (Same f) -V_{s2} Higher Freq. (f₂ > f₁)

3. Examples

Example 1: Work Function and Threshold Analysis

Problem: The threshold frequency for a particular metal is $4.0 \times 10^{14} \text{ Hz}$. Light of frequency $6.5 \times 10^{14} \text{ Hz}$ is shone onto the surface.
(a) Calculate the work function of the metal in eV.
(b) Calculate the maximum kinetic energy of the emitted photoelectrons in eV.

Solution:

Part (a): The work function is related to the threshold frequency by $\Phi = hf_0$. $$\Phi = (6.63 \times 10^{-34} \text{ J s}) \times (4.0 \times 10^{14} \text{ Hz}) = 2.652 \times 10^{-19} \text{ J}$$ Convert Joules to electronvolts (eV): $$\Phi = \frac{2.652 \times 10^{-19} \text{ J}}{1.60 \times 10^{-19} \text{ J/eV}} \approx \mathbf{1.66 \text{ eV}}$$

Part (b): Determine the energy of the incident photon: $$E_{\text{photon}} = hf = (6.63 \times 10^{-34}) \times (6.5 \times 10^{14}) = 4.3095 \times 10^{-19} \text{ J}$$ Convert to eV: $$E_{\text{photon}} = \frac{4.3095 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 2.69 \text{ eV}$$ Use the photoelectric equation: $$E_{k \text{max}} = E_{\text{photon}} - \Phi = 2.69 \text{ eV} - 1.66 \text{ eV} = \mathbf{1.03 \text{ eV}}$$

Example 2: Stopping Potential and Wavelength Changes

Problem: Monochromatic light of wavelength $\lambda_1$ is incident on the surface of a metal. The stopping potential for this light is $V_1$. When another monochromatic light of wavelength $\lambda_2$ is incident on the same surface, the stopping potential is $V_2$. Derive an expression for the ratio $\dfrac{hc}{e}$ in terms of $V_1, V_2, \lambda_1,$ and $\lambda_2$.

Solution:

Step 1: Set up the photoelectric equation for both scenarios:
For the first light source: $$\frac{hc}{\lambda_1} = \Phi + eV_1 \implies eV_1 = \frac{hc}{\lambda_1} - \Phi \quad \text{--- (Equation 1)}$$ For the second light source: $$\frac{hc}{\lambda_2} = \Phi + eV_2 \implies eV_2 = \frac{hc}{\lambda_2} - \Phi \quad \text{--- (Equation 2)}$$

Step 2: Subtract Equation 1 from Equation 2 to eliminate the work function $\Phi$: $$eV_2 - eV_1 = \left(\frac{hc}{\lambda_2} - \Phi\right) - \left(\frac{hc}{\lambda_1} - \Phi\right)$$ $$e(V_2 - V_1) = hc \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$$

Step 3: Rearrange to isolate $\dfrac{hc}{e}$: $$\mathbf{\frac{hc}{e} = \frac{V_2 - V_1}{\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right)}}$$ This technique demonstrates how the Planck constant can be determined experimentally using stopping potentials and varying light frequencies.

Example 3: Stopping Potential

Problem: In a photoelectric experiment, the stopping potential $V_s$ is measured for various frequencies $f$ of incident light. The results are plotted on a graph of $V_s$ (y-axis) against $f$ (x-axis), yielding a straight line. The line intersects the x-axis at $f_0 = 4.60 \times 10^{14} \text{ Hz}$. Another point on the line is recorded at $f = 10.0 \times 10^{14} \text{ Hz}$ with a stopping potential of $V_s = 2.24 \text{ V}$.
(a) Using only the data from this graph, calculate the experimental value for Planck's constant, $h$.
(b) Determine the work function of the metal in eV.

Solution:

Part (a): Calculate $h$:
Start with the photoelectric equation: $hf = \Phi + E_{k\text{max}}$. Since $E_{k\text{max}} = eV_s$, we can write $$ eV_s = hf - \Phi \implies V_s = \left(\frac{h}{e}\right)f - \frac{\Phi}{e} $$ This matches the equation of a line $y = mx + c$. The slope of a $V_s$-vs-$f$ graph is equal to $\dfrac{h}{e}$. $$ \begin{aligned} \text{Slope} &= \frac{\Delta V_s}{\Delta f} = \frac{2.24 \text{ V} - 0 \text{ V}}{(10.0 \times 10^{14} \text{ Hz}) - (4.60 \times 10^{14} \text{ Hz})} \\ &= \frac{2.24}{5.40 \times 10^{14}} = 4.148 \times 10^{-15} \text{ V s} \end{aligned} $$ Now, multiply the gradient by the elementary charge $e$ to find $h$: $$h = \text{Slope} \times e = (4.148 \times 10^{-15}) \times (1.60 \times 10^{-19}) = \mathbf{6.64 \times 10^{-34} \text{ J s}}$$

Part (b): Calculate $\Phi$:
The x-intercept is the threshold frequency $f_0 = 4.60 \times 10^{14} \text{ Hz}$. At this point, kinetic energy is zero, so $hf_0 = \Phi$. $$\Phi = (6.64 \times 10^{-34} \text{ J s}) \times (4.60 \times 10^{14} \text{ Hz}) = 3.05 \times 10^{-19} \text{ J}$$ Convert to eV: $$ \Phi = \frac{3.05 \times 10^{-19}}{1.60 \times 10^{-19}} = \mathbf{1.91 \text{ eV}} $$

Example 4: Quantum Yield and Saturation Photocurrent

Problem: A laser emits monochromatic light of wavelength $410 \text{ nm}$ with a power output of $5.0 \text{ mW}$. The beam is directed onto a photosensitive surface. The quantum efficiency (the probability that an incident photon successfully ejects a photoelectron) is $4.0\%$. Calculate the maximum saturation photoelectric current (in microamperes, $\mu\text{A}$) that can be measured in the circuit.

Solution:

Step 1: Find the energy of a single photon: $$E_{ph} = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{410 \times 10^{-9}} = 4.85 \times 10^{-19} \text{ J}$$

Step 2: Find the number of incident photons per second ($N_p$): Power is energy per second ($5.0 \text{ mW} = 5.0 \times 10^{-3} \text{ J/s}$). $$N_p = \frac{\text{Total Power}}{\text{Energy per photon}} = \frac{5.0 \times 10^{-3}}{4.85 \times 10^{-19}} = 1.03 \times 10^{16} \text{ photons/s}$$

Step 3: Apply quantum efficiency to find ejected electrons per second ($N_e$): Only $4.0\%$ of photons eject an electron. $$N_e = 0.04 \times (1.03 \times 10^{16}) = 4.12 \times 10^{14} \text{ electrons/s}$$

Step 4: Calculate the total current ($I$): Current is the total charge flowing per second ($I = N_e \times e$). $$I = (4.12 \times 10^{14}) \times (1.60 \times 10^{-19} \text{ C}) = 6.59 \times 10^{-5} \text{ A} = \mathbf{65.9 \ \mu\text{A}}$$