Photoelectric Effect: Equations, Calculations, and Applications

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Photoelectric Effect

Introduction to the Photoelectric Effect

The photoelectric effect is the phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation of sufficient frequency. This effect provided crucial evidence for the quantum nature of light and led to the development of quantum mechanics.

Key Concept: Electrons are ejected only if the incident light has a frequency above a certain threshold, regardless of intensity.
Work Function ($ \phi $): The minimum energy required to remove an electron from the surface of a material.
Threshold Frequency ($ f_c $): The minimum frequency of light needed to emit electrons from a material.

Photoelectric Effect Equation

The maximum kinetic energy of emitted electrons is given by:

$KE_{\text{max}} = E_{\text{photon}} - \phi = hf - \phi$

$ KE_{\text{max}} $: Maximum kinetic energy of ejected electrons
$ E_{\text{photon}} = hf $: Energy of the incident photon
$ h $: Planck's constant ($ 6.63 \times 10^{-34} $ J·s)
$ f $: Frequency of incident light
$ \phi $: Work function of the material

Cutoff Wavelength and Frequency

The cutoff wavelength ($ \lambda_c $) is the longest wavelength (lowest energy) that can cause electron emission. It is related to the work function by:

$\lambda_c = \frac{hc}{\phi}$

The cutoff frequency ($ f_c $) is the minimum frequency required for electron emission:

$f_c = \frac{c}{\lambda_c} = \frac{\phi}{h}$

$ c $: Speed of light ($ 3.00 \times 10^8 $ m/s)

Example Calculations

1. Calculating the Cutoff Wavelength

Given $ \phi = 2.24 $ eV, the cutoff wavelength is:

$\lambda_c = \frac{(6.63 \times 10^{-34} \ \text{J·s})(3.00 \times 10^8 \ \text{m/s})}{2.24 \ \text{eV}} \left( \frac{1 \ \text{eV}}{1.60 \times 10^{-19} \ \text{J}} \right) = 5.55 \times 10^{-7} \ \text{m} = 555 \ \text{nm}$

Calculation of cutoff wavelength

2. Calculating the Cutoff Frequency

The lowest frequency of light that will free electrons is:

$f_c = \frac{c}{\lambda_c} = \frac{3.00 \times 10^8 \ \text{m/s}}{2.88 \times 10^{-7} \ \text{m}} = 1.04 \times 10^{15} \ \text{Hz}$

Calculation of cutoff frequency

3. Calculating the Work Function from Wavelength

Given a wavelength and photon energy, the work function can be found as:

$\phi = \frac{(6.63 \times 10^{-34} \ \text{J·s})(3.00 \times 10^8 \ \text{m/s})}{350 \times 10^{-9} \ \text{m}} \left( \frac{1 \ \text{eV}}{1.60 \times 10^{-19} \ \text{J}} \right) - 1.31 \ \text{eV} = 2.24 \ \text{eV}$

Calculation of work function

4. Photoelectric Effect Equation at Cutoff

At the cutoff wavelength, $ KE_{\text{max}} = 0 $, so:

$KE_{\text{max}} = \frac{hc}{\lambda} - \phi$

Photoelectric effect equation at cutoff

5. Example: Cutoff Wavelength for a Given Work Function

Given $ \phi = 4.31 $ eV:

$\lambda_c = \frac{(6.63 \times 10^{-34} \ \text{J·s})(3.00 \times 10^8 \ \text{m/s})}{4.31 \ \text{eV}} \left( \frac{1 \ \text{eV}}{1.60 \times 10^{-19} \ \text{J}} \right) = 2.88 \times 10^{-7} \ \text{m} = 288 \ \text{nm}$

Cutoff wavelength calculation

6. Maximum Kinetic Energy of Ejected Electrons

If the photon energy is $ E_{\text{photon}} = 5.50 $ eV and $ \phi = 4.31 $ eV:

$KE_{\text{max}} = E_{\text{photon}} - \phi = 5.50 \ \text{eV} - 4.31 \ \text{eV} = 1.19 \ \text{eV}$

Maximum kinetic energy calculation

7. Cutoff Frequency for a Given Work Function

For $ \phi = 6.35 $ eV:

$f_c = \frac{\phi}{h} = \frac{6.35 \ \text{eV}}{6.63 \times 10^{-34} \ \text{J·s}} \left(1.60 \times 10^{-19} \ \text{J/eV}\right) = 1.53 \times 10^{15} \ \text{Hz}$

Cutoff frequency calculation

8. Wavelength from Frequency

Given $ f_c = 1.53 \times 10^{15} $ Hz:

$\lambda_c = \frac{c}{f_c} = \frac{3.00 \times 10^8 \ \text{m/s}}{1.53 \times 10^{15} \ \text{Hz}} = 1.96 \times 10^{-7} \ \text{m} = 196 \ \text{nm}$

Wavelength from frequency calculation

Stopping Potential

The stopping potential ($ V_s $) is the minimum voltage needed to stop the most energetic photoelectrons:

$eV_s = KE_{\text{max}} \implies V_s = \frac{KE_{\text{max}}}{e}$

$ e $: Elementary charge ($ 1.60 \times 10^{-19} $ C)

Example: If $ KE_{\text{max}} = 2.15 $ eV, then $ V_s = 2.15 $ V.

Stopping potential calculation

Conservation of Energy and Maximum Kinetic Energy

By conservation of energy, the maximum kinetic energy equals $ eV_s $, the highest potential energy achieved by the photoelectrons.

Example: If $ V_s = 2.50 $ V, then $ KE_{\text{max}} = 2.50 $ eV.

Maximum kinetic energy from stopping potential

Calculating Electron Velocity from Kinetic Energy

The maximum velocity of ejected electrons can be found from kinetic energy:

$KE = \frac{1}{2} m_e v_{\text{max}}^2$

$v_{\text{max}} = \sqrt{\frac{2 KE}{m_e}}$

$ m_e $: Electron mass ($ 9.11 \times 10^{-31} $ kg)

Example: For $ KE = 4.00 \times 10^{-19} $ J, $ v_{\text{max}} = 9.37 \times 10^5 $ m/s.

Electron velocity calculation

Photon Energy from Wavelength

The energy of a photon is related to its wavelength by:

$E_{\text{photon}} = \frac{hc}{\lambda}$

Example: For $ \lambda = 400 $ nm, $ E_{\text{photon}} = 3.11 $ eV.

Photon energy from wavelength

Wavelength for a Given Kinetic Energy and Work Function

By conservation of energy, $ KE_{\text{max}} = hf - \phi $. Substitute $ f = c/\lambda $ and solve for $ \lambda $:

$\lambda = \frac{hc}{KE_{\text{max}} + \phi}$

Example: For $ KE_{\text{max}} = 2.50 $ eV and $ \phi = 4.73 $ eV, $ \lambda = 172 $ nm.

Wavelength for given KE and work function

Work Function from Wavelength and Kinetic Energy

The work function can also be found from the cutoff wavelength and kinetic energy:

$\phi = \frac{hc}{\lambda} - KE_{\text{max}}$

Example: For $ \lambda = 546.1 $ nm and $ KE_{\text{max}} = 0.376 $ eV, $ \phi = 1.90 $ eV.

Work function from wavelength and KE

Stopping Potential from Kinetic Energy

The stopping potential can be calculated as:

$e(\Delta V_s) = KE_{\text{max}}$

Example: For $ KE_{\text{max}} = 0.216 $ eV, $ \Delta V_s = 0.216 $ V.

Stopping potential from kinetic energy

Graphical Representation of the Photoelectric Effect

The relationship between the maximum kinetic energy of photoelectrons and the frequency of incident light is linear above the threshold frequency. The slope of the line is Planck's constant, and the intercept on the frequency axis gives the threshold frequency.

Graph of KE_max vs frequency Another graph of KE_max vs frequency

Summary Table: Key Photoelectric Effect Equations

Quantity	Equation	Description
Maximum Kinetic Energy	$KE_{\text{max}} = hf - \phi$	Energy of ejected electrons
Cutoff Frequency	$f_c = \frac{\phi}{h}$	Minimum frequency for emission
Cutoff Wavelength	$\lambda_c = \frac{hc}{\phi}$	Maximum wavelength for emission
Stopping Potential	$V_s = \frac{KE_{\text{max}}}{e}$	Voltage to stop electrons
Photon Energy	$E_{\text{photon}} = hf = \frac{hc}{\lambda}$	Energy of incident photon

Additional info: The photoelectric effect is a foundational experiment in modern physics, confirming the quantization of light and leading to the development of quantum mechanics. The equations above are essential for solving problems related to the emission of electrons from metals under light exposure, and for understanding the energy relationships involved.