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Mechanism of the Photoelectric Effect

## Article: Mechanism of the Photoelectric Effect

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it absorbs electromagnetic radiation, such as light. This concept was first observed in 1887 by Heinrich Hertz and later explained by Albert Einstein, earning him the Nobel Prize in Physics in 1921.

### Understanding the Mechanism

The photoelectric effect can be observed when light or photons strike a material, typically a metal, with sufficient energy. The energy of a photon is given by the equation:

\[ E = h \nu \]

where \( E \) is the energy, \( h \) is Planck’s constant, and \( \nu \) (nu) is the frequency of the light.

For the photoelectric effect to occur, the energy of the photons must be greater than the work function (\( \phi \)) of the material, which is the minimum energy required to remove an electron from the surface. If the photon energy exceeds the work function, the excess energy is converted into the kinetic energy (\( K.E. \)) of the emitted electron, as described by the equation:

\[ K.E. = h \nu – \phi \]

This equation is known as the photoelectric equation or Einstein’s photoelectric equation.

### Key Points of the Photoelectric Effect

1. **Threshold Frequency**: There is a minimum frequency, known as the threshold frequency (\( \nu_0 \)), below which no electrons are emitted, regardless of the intensity of the light.

2. **Intensity and Number of Electrons**: The intensity of the light affects the number of electrons emitted but does not affect their kinetic energy.

3. **Kinetic Energy and Frequency**: The kinetic energy of the emitted electrons is directly related to the frequency of the incident light and not the intensity.

4. **Instantaneous Effect**: The emission of electrons occurs almost instantaneously (within nanoseconds) after the light is absorbed, indicating a direct energy transfer from photons to electrons.

### Applications

The photoelectric effect has numerous applications, including the development of photoelectric cells, which are used to convert light into electricity in solar panels, light meters used in photography, and the detection of light in electronic devices like television remote controls.

### Problems and Solutions About the Mechanism of the Photoelectric Effect

1. **Problem**: Calculate the energy of a photon with a frequency of \(6.0 \times 10^{14}\) Hz. (Take Planck’s constant, \(h\), as \(6.626 \times 10^{-34}\) Js)

**Solution**:

\[ E = h \nu \]

\[ E = (6.626 \times 10^{-34} \, \text{Js})(6.0 \times 10^{14} \, \text{Hz}) \]

\[ E = 3.976 \times 10^{-19} \, \text{J} \]

2. **Problem**: If the energy of a photon is \(3.2 \times 10^{-19}\) J, what is its frequency? (Take Planck’s constant, \(h\), as \(6.626 \times 10^{-34}\) Js)

**Solution**:

\[ E = h \nu \]

\[ \nu = \frac{E}{h} \]

\[ \nu = \frac{3.2 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \]

\[ \nu = 4.83 \times 10^{14} \, \text{Hz} \]

3. **Problem**: Find the threshold frequency for a metal whose work function is \(2.5 \times 10^{-19}\) J. (Use Planck’s constant, \(h\), as \(6.626 \times 10^{-34}\) Js)

**Solution**:

\[ \phi = h \nu_0 \]

\[ \nu_0 = \frac{\phi}{h} \]

\[ \nu_0 = \frac{2.5 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \]

\[ \nu_0 = 3.77 \times 10^{14} \, \text{Hz} \]

4. **Problem**: A photon of frequency \(7.5 \times 10^{14}\) Hz strikes a metal surface with a work function of \(2.0 \times 10^{-19}\) J. Calculate the kinetic energy of the emitted electron.

**Solution**:

\[ K.E. = h \nu – \phi \]

\[ K.E. = (6.626 \times 10^{-34} \, \text{Js})(7.5 \times 10^{14} \, \text{Hz}) – 2.0 \times 10^{-19} \, \text{J} \]

\[ K.E. = 4.97 \times 10^{-19} \, \text{J} – 2.0 \times 10^{-19} \, \text{J} \]

\[ K.E. = 2.97 \times 10^{-19} \, \text{J} \]

5. **Problem**: What is the maximum wavelength of light that can eject electrons from a metal with a work function of \(3.1 \times 10^{-19}\) J?

**Solution**:

Use the relation \( c = \lambda \nu \) to find the wavelength \( \lambda \), where \( c \) is the speed of light:

\[ \phi = h \frac{c}{\lambda} \]

\[ \lambda = \frac{h c}{\phi} \]

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^{8} \, \text{m/s})}{3.1 \times 10^{-19} \, \text{J}} \]

\[ \lambda = 6.42 \times 10^{-7} \, \text{m} \]

\[ \lambda = 642 \, \text{nm} \]

6. **Problem**: Determine the stopping potential needed to stop electrons with a kinetic energy of \(4.0 \times 10^{-19}\) J.

**Solution**:

The stopping potential \( V \) can be found by equating the kinetic energy to the electric potential energy ( \( eV \), where \( e \) is the charge of an electron):

\[ K.E. = eV \]

Using the elementary charge \( e = 1.602 \times 10^{-19} \, \text{C} \):

\[ V = \frac{K.E.}{e} \]

\[ V = \frac{4.0 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{C}} \]

\[ V = 2.50 \, \text{V} \]

7. **Problem**: A metal with a threshold frequency of \(4.0 \times 10^{14}\) Hz is illuminated with light of frequency \(5.0 \times 10^{14}\) Hz. What is the kinetic energy of the emitted electrons?

**Solution**:

\[ K.E. = h (\nu – \nu_0) \]

\[ K.E. = (6.626 \times 10^{-34} \, \text{Js})(5.0 \times 10^{14} \, \text{Hz} – 4.0 \times 10^{14} \, \text{Hz}) \]

\[ K.E. = 6.626 \times 10^{-34} \, \text{Js} \times 1.0 \times 10^{14} \, \text{Hz} \]

\[ K.E. = 6.626 \times 10^{-20} \, \text{J} \]

8. **Problem**: Calculate the work function of a metal if the stopping potential for electrons ejected by 200 nm light is 1.2 V.

**Solution**:

Convert the wavelength \( \lambda \) to frequency \( \nu \) using \( c = \lambda \nu \):

\[ \nu = \frac{c}{\lambda} \]

\[ \nu = \frac{3.00 \times 10^{8} \, \text{m/s}}{200 \times 10^{-9} \, \text{m}} \]

\[ \nu = 1.5 \times 10^{15} \, \text{Hz} \]

Now find the photon energy \( E \):

\[ E = h \nu \]

\[ E = (6.626 \times 10^{-34} \, \text{Js})(1.5 \times 10^{15} \, \text{Hz}) \]

\[ E = 9.939 \times 10^{-19} \, \text{J} \]

The work function \( \phi \) can be determined by equating the stopping potential and kinetic energy:

\[ eV = E – \phi \]

\[ \phi = E – eV \]

\[ \phi = 9.939 \times 10^{-19} \, \text{J} – (1.602 \times 10^{-19} \, \text{C})(1.2 \, \text{V}) \]

\[ \phi = 9.939 \times 10^{-19} \, \text{J} – 1.922 \times 10^{-19} \, \text{J} \]

\[ \phi = 8.017 \times 10^{-19} \, \text{J} \]

9. **Problem**: A light of frequency \(8.2 \times 10^{14}\) Hz is incident on a metal surface with a work function of \(4.5 \times 10^{-19}\) J. Calculate the maximum speed of the emitted electrons.

**Solution**:

\[ K.E. = h \nu – \phi \]

\[ K.E. = (6.626 \times 10^{-34} \, \text{Js})(8.2 \times 10^{14} \, \text{Hz}) – 4.5 \times 10^{-19} \, \text{J} \]

\[ K.E. = 5.4332 \times 10^{-19} \, \text{J} – 4.5 \times 10^{-19} \, \text{J} \]

\[ K.E. = 9.332 \times 10^{-20} \, \text{J} \]

Use the kinetic energy formula \( K.E. = \frac{1}{2}mv^2 \) to find the velocity \( v \), and assume \( m \) is the mass of an electron (\( m = 9.109 \times 10^{-31} \, \text{kg} \) ):

\[ v = \sqrt{\frac{2 K.E.}{m}} \]

\[ v = \sqrt{\frac{2 \times 9.332 \times 10^{-20} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} \]

\[ v = \sqrt{\frac{1.8664 \times 10^{-19} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} \]

\[ v \approx 6.045 \times 10^{5} \, \text{m/s} \]

10. **Problem**: If the stopping potential for a certain metal is 0.5 V when illuminated with light of frequency \(6.5 \times 10^{14}\) Hz, what is the work function of the metal?

**Solution**:

\[ eV = h \nu – \phi \]

\[ \phi = h \nu – eV \]

\[ \phi = (6.626 \times 10^{-34} \, \text{Js})(6.5 \times 10^{14} \, \text{Hz}) – (1.602 \times 10^{-19} \, \text{C})(0.5 \, \text{V}) \]

\[ \phi = 4.307 \times 10^{-19} \, \text{J} – 8.01 \times 10^{-20} \, \text{J} \]

\[ \phi = 3.507 \times 10^{-19} \, \text{J} \]

11. **Problem**: A metal has a work function of \(2.8 \times 10^{-19}\) J. If the kinetic energy of the ejected electrons is \(6.0 \times 10^{-19}\) J, what is the frequency of the incident light?

**Solution**:

\[ K.E. = h \nu – \phi \]

\[ \nu = \frac{K.E. + \phi}{h} \]

\[ \nu = \frac{6.0 \times 10^{-19} \, \text{J} + 2.8 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \]

\[ \nu = \frac{8.8 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \]

\[ \nu \approx 1.33 \times 10^{15} \, \text{Hz} \]

12. **Problem**: What is the threshold wavelength for a metal if its threshold frequency is \(5.5 \times 10^{14}\) Hz?

**Solution**:

\[ \lambda = \frac{c}{\nu_0} \]

\[ \lambda = \frac{3.00 \times 10^{8} \, \text{m/s}}{5.5 \times 10^{14} \, \text{Hz}} \]

\[ \lambda \approx 545.45 \, \text{nm} \]

13. **Problem**: If electrons are being emitted with no kinetic energy, what is the frequency of the incident light on a metal with a work function of \(3.3 \times 10^{-19}\) J?

**Solution**:

If the electrons have no kinetic energy, the frequency of the light must be the threshold frequency:

\[ \phi = h \nu_0 \]

\[ \nu_0 = \frac{\phi}{h} \]

\[ \nu_0 = \frac{3.3 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \]

\[ \nu_0 \approx 4.98 \times 10^{14} \, \text{Hz} \]

14. **Problem**: A metal with threshold frequency of \(4.5 \times 10^{14}\) Hz is exposed to light of wavelength 300 nm. Determine if photoelectrons will be emitted.

**Solution**:

Convert the wavelength to frequency:

\[ \nu = \frac{c}{\lambda} \]

\[ \nu = \frac{3.00 \times 10^{8} \, \text{m/s}}{300 \times 10^{-9} \, \text{m}} \]

\[ \nu = 1.00 \times 10^{15} \, \text{Hz} \]

Since \( \nu > \nu_0 \) (1.00 x \(10^{15}\) Hz > 4.5 x \(10^{14}\) Hz), photoelectrons will be emitted.

15. **Problem**: What must be the minimum voltage applied to stop the photoelectrons if the light of wavelength 250 nm is incident on a metal with the work function of \(2.25 \times 10^{-19}\) J?

**Solution**:

Convert the wavelength to frequency and then calculate the photon energy:

\[ \nu = \frac{c}{\lambda} \]

\[ \nu = \frac{3.00 \times 10^{8} \, \text{m/s}}{250 \times 10^{-9} \, \text{m}} \]

\[ \nu = 1.2 \times 10^{15} \, \text{Hz} \]

\[ E = h \nu \]

\[ E = (6.626 \times 10^{-34} \, \text{Js})(1.2 \times 10^{15} \, \text{Hz}) \]

\[ E = 7.9512 \times 10^{-19} \, \text{J} \]

Calculate the kinetic energy and then the stopping potential:

\[ K.E. = E – \phi \]

\[ K.E. = 7.9512 \times 10^{-19} \, \text{J} – 2.25 \times 10^{-19} \, \text{J} \]

\[ K.E. = 5.7012 \times 10^{-19} \, \text{J} \]

\[ V = \frac{K.E.}{e} \]

\[ V = \frac{5.7012 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{C}} \]

\[ V \approx 3.56 \, \text{V} \]

16. **Problem**: How much kinetic energy would an electron have if it is ejected by a photon with a wavelength of 400 nm from a metal with a work function of \(3.6 \times 10^{-19}\) J?

**Solution**:

Convert wavelength to frequency and calculate the photon energy:

\[ \nu = \frac{c}{\lambda} \]

\[ \nu = \frac{3.00 \times 10^{8} \, \text{m/s}}{400 \times 10^{-9} \, \text{m}} \]

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