Photons and the Photoelectric Effect Study Guide

Quantum Theory of Radiation and Photons

According to the quantum theory of radiation, the energy emitted from a body is not continuous but is emitted in separate packets of energy called quantum of energy. The energy carried by radiation is quantized. Each discrete bundle or packet of radiation that carries a certain amount of energy is called a photon.

Characteristics of Photons

• Charge: Photons are chargeless, which means they are not deflected by electric or magnetic fields.
• Speed: Photons travel in a straight line with the speed of light in a vacuum ($c = 3 \times 10^8 \, \text{m/s}$).
• Momentum: The momentum of a photon is given by:   $P = \frac{h}{\lambda}$
• Energy: The energy of each photon is calculated as:   $E = hf$
• Force and Pressure: A photon exerts force and pressure when it strikes a surface.

Fundamental Terms in Photoelectricity

Photoelectric Effect

The phenomenon of emission or ejection of electrons from a metal surface when radiation of a suitable frequency is incident upon it is called the photoelectric effect.

Photoelectron

The electrons which are emitted from a metallic surface during the photoelectric effect are called photoelectrons.

Photoelectric Current

The current flow through a metallic surface due to the photoelectric effect is called photoelectric current ($I_p$). In the photoelectric effect, if the mass of a photoelectron is $m$ and it is moving with a certain velocity $v$, then:

$K.E. = \frac{1}{2}mv^2$

Work Function

The minimum energy required just to eject a photoelectron from a metal surface is called the work function. It is denoted by $\Phi$ or $W_0$ and is given by:

$\Phi = hf_0$

Where:

• $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ or approximate value $6.6 \times 10^{-34} \, \text{J} \cdot \text{s}$).
• $f_0$ is the threshold frequency.

Using the relationship $c = \lambda_0 f_0$, the work function can also be expressed as:

$\Phi = \frac{hc}{\lambda_0}$

Where $\lambda_0$ is the threshold wavelength. The work function depends upon the nature of the materials and does not depend upon the intensity of the radiation falling on it.

Threshold Threshold Parameters

Threshold Frequency ($f_0$)

• The minimum frequency of incident radiation below which the photoelectric effect does not happen is called the threshold frequency.
• It is also known as the cut-off frequency.
• It is denoted by $f_0$ and is given by $f_0 = \frac{c}{\lambda_0}$.
• From the definition of the work function, $f_0 = \frac{\Phi}{h}$.
• It depends upon the nature of the material.

Threshold Wavelength ($\lambda_0$)

• The maximum wavelength of incident radiation above which no photoelectric effect or emission happens is called the threshold wavelength.
• It is denoted by $\lambda_0$.
• It depends upon the nature of the material.

Critical Conditions for Emission

• If the frequency of incident radiation $f < f_0$, there is no photoelectric effect regardless of intensity.
• If radiation is incident on a metallic surface with frequency $f > f_0$, the electron acquires a maximum velocity:

$K.E._{max} = \frac{1}{2}mv_{max}^2$

Stopping Potential

The minimum value of negative potential applied to the anode which can just stop the photoelectrons from the metal surface (making the photoelectric current zero) is called the stopping potential ($V_0$). If the photoelectric current becomes zero, the work done by the potential corresponds to the maximum kinetic energy:

$eV_0 = \frac{1}{2}mv^2_{max}$

Quantitative Exercises and Examples

Light Source and Photon Count

A $75 \, \text{W}$ light source consumes $75 \, \text{Joules}$ of electrical energy per second. Assuming all energy is emitted as light of $\lambda = 600 \, \text{nm}$:

1. Frequency of emitted light:$f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{600 \times 10^{-9} \, \text{m}} = 5 \times 10^{14} \, \text{Hz}$

2. Number of photons per second ($N$):$P = \frac{n}{t}hf = Nhf$$75 = N \times (6.626 \times 10^{-34}) \times (5 \times 10^{14})$$N = \frac{75}{3.313 \times 10^{-19}} = 2.263 \times 10^{20} \, \text{photons/sec}$

Green Light Properties

A photon of green light has a wavelength of $520 \, \text{nm}$.

• Frequency: $f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{520 \times 10^{-9} \, \text{m}} = 5.76 \times 10^{14} \, \text{Hz}$
• Momentum: $P = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}}{520 \times 10^{-9} \, \text{m}} = 1.27 \times 10^{-27} \, \text{kg} \cdot \text{m/s}$
• Energy (Joules): $E = hf = (6.626 \times 10^{-34}) \times (5.76 \times 10^{14}) = 3.82 \times 10^{-19} \, \text{J}$
• Energy (eV): $E = \frac{3.82 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 2.39 \, \text{eV}$

Einstein’s Photoelectric Equation

When a photon of frequency $f$ is incident on a metal surface, its energy ($E = hf$) is completely transferred to a free electron. This energy is used in two ways:

1. A certain amount ($\Phi$) is used to eject the electron from the surface.
2. The remaining energy is converted into the kinetic energy of the electron.

According to the conservation of energy: $E = \Phi + K.E._{max}$$hf = hf_0 + \frac{1}{2}mv^2_{max}$

Alternative Forms

• In terms of frequency: $\frac{1}{2}mv^2_{max} = h(f - f_0)$
• In terms of wavelength: $\frac{1}{2}mv^2_{max} = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$

Observational Cases

• Case 1 ($f > f_0$): Photoelectric effect is possible.
• Case 2 ($f < f_0$): Photoelectric effect is not possible.
• Kinetic Energy: Depends directly on the frequency of incident radiation, not on the intensity.
• Rate of Emission: Depends directly on the intensity of radiation, not on the frequency.

Experimental Study of Photoelectric Effect

Experimental Arrangement

The setup consists of an evacuated glass or quartz tube containing two electrodes: an anode ($A$) and a cathode ($C$). The cathode is made of photo-sensitive alkali metal. The electrodes are connected to a potential divider to change the potential difference. Photoelectric current ($I_p$) is measured by a milliammeter ($mA$) and potential difference by a voltmeter ($V$).

When light of suitable frequency enters the window ($w$) and hits the cathode, electrons are emitted and accelerated toward the anode by a positive potential. This produced current flows in the external circuit.

Characteristics Derived Experimentally

1. Time Lag: The flow of current reaches a steady point in about $10^{-9} \, \text{s}$ from the start of irradiation, independent of intensity.
2. Intensity: Photoelectric current ($I_p$) is directly proportional to intensity ($I$) of incident radiation ($I_p \propto I$).
3. Potential:
   • As positive anode potential increases, current increases until it reaches a saturation point, after which it remains constant.
   • If negative potential (retarding potential) is applied, current decreases. The specific negative potential where current becomes zero is the stopping potential ($V_0$).
   • $V_0$ is independent of the intensity of light but directly proportional to the frequency of incident light.

Millikan’s Verification of Einstein's Equation

Millikan used an evacuated glass chamber with a rotating wheel containing cylindrical blocks of alkali metals (Sodium, Potassium, Lithium). A knife was used to remove the oxide layer from the metal surfaces to ensure cleanliness.

By measuring stopping potentials for different frequencies of radiation and plotting a graph of $V_0$ vs. $f$, a straight line is obtained.

Calculation of Planck's Constant ($h$): From stopping potential condition: $eV_0 = \frac{1}{2}mv^2_{max}$ From Einstein's eqn: $hf = hf_0 + eV_0$$V_0 = \left(\frac{h}{e}\right)f - \left(\frac{hf_0}{e}\right)$

This represents a straight line ($y = mx + c$) where:

• Slope ($m$): $\frac{h}{e}$
• y-intercept: $-\frac{hf_0}{e}$

By finding the slope ($\tan(\theta) = \frac{\Delta V_0}{\Delta f}$) and knowing the charge of an electron ($e$), Planck's constant is calculated ($h = e \times \text{slope}$). Millikan found $h \approx 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$, verifying Einstein's theory.

Practical Applications and Short Q&A

Applications

• Photoelectric cells.
• Automatic photographic cameras.
• Electronic devices like television and computers.
• Sound reproduction in cinematography.

Conceptual Questions

• Why are alkali metals suited for emission? They have very low work function values, meaning less energy is needed to eject electrons.
• Is it harder to remove electrons from Copper or Sodium? Copper is more difficult because it has a higher work function than Sodium.
• Can one photon eject multiple electrons? No. A photon acts as a single particle and its energy cannot be shared. One photon interacts with one electron.
• Visible light emission: Alkali metals show the effect with visible light because the photon energy of visible light is sufficient to overcome their low work functions.

Detailed Problem Solving

Maximum Kinetic Energy Calculation

Sodium Work function $\Phi = 2.88 \times 10^{-19} \, \text{J}$, Mercury light frequency $f = 4.7 \times 10^{14} \, \text{Hz}$. $E = hf = (6.62 \times 10^{-34}) \times (4.7 \times 10^{14}) = 3.11 \times 10^{-19} \, \text{J}$$K.E._{max} = E - \Phi = 3.11 \times 10^{-19} - 2.88 \times 10^{-19} = 2.31 \times 10^{-20} \, \text{J}$

Cesium Surface Example

Work function $\Phi = 1.25 \, \text{eV}$. Light wavelength $\lambda = 4 \times 10^{-7} \, \text{m}$.

1. Threshold Wavelength:$\lambda_0 = \frac{hc}{\Phi} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.25 \times 1.6 \times 10^{-19}} = 9.9 \times 10^{-7} \, \text{m}$
2. Maximum Velocity:$K.E._{max} = \frac{hc}{\lambda} - \Phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} - (1.25 \times 1.6 \times 10^{-19})$$K.E._{max} = 4.95 \times 10^{-19} - 2.0 \times 10^{-19} = 2.95 \times 10^{-19} \, \text{J}$$v_{max} = \sqrt{\frac{2 \times K.E._{max}}{m}} = \sqrt{\frac{2 \times 2.95 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 8.08 \times 10^5 \, \text{m/s}$

Solving for Threshold Frequency

If $K.E._{max} = 1.6 \times 10^{-19} \, \text{J}$ at $f = 7.5 \times 10^{14} \, \text{Hz}$, assume $h = 6.62 \times 10^{-34} \, \text{J} \cdot \text{s}$. $1.6 \times 10^{-19} = 6.62 \times 10^{-34} \times (7.5 \times 10^{14} - f_0)$$2.416 \times 10^{14} = 7.5 \times 10^{14} - f_0$$f_0 = 5.084 \times 10^{14} \, \text{Hz}$

Multiple Choice Review

1. Which doesn't explain wave theory? Photo-electric effect.
2. Energy of photon representation: Energy cannot be represented by $hv \lambda$ (it is $hf$ or $hc/\lambda$).
3. Increase in intensity effects: Increases photoelectric current.
4. Mass of photon: Rest mass is $0$, mass in motion is $hf/c^2$.
5. Principle of effect: Based on the conservation of Energy.