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PHYSICS, M7 2025 HSC 25

A student conducts an experiment to determine the work function of potassium.

The following diagram depicts the experimental setup used, where light of varying frequency is incident on a potassium electrode inside an evacuated tube.
 

For each frequency of light tested, the voltage in the circuit is varied, and the minimum voltage (called the stopping voltage) required to bring the current in the circuit to zero is recorded.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{Frequency of incident light} \ \ & \quad \quad \quad \textit{Stopping voltage} \quad \quad \quad \\
\left(\times 10^{15} \ \text{Hz}\right) \rule[-1ex]{0pt}{0pt}& \text{(V)}\\
\hline
\rule{0pt}{2.5ex}0.9 \rule[-1ex]{0pt}{0pt}& 1.5 \\
\hline
\rule{0pt}{2.5ex}1.1 \rule[-1ex]{0pt}{0pt}& 2.0 \\
\hline
\rule{0pt}{2.5ex}1.2 \rule[-1ex]{0pt}{0pt}& 2.5 \\
\hline
\rule{0pt}{2.5ex}1.3 \rule[-1ex]{0pt}{0pt}& 3.0 \\
\hline
\rule{0pt}{2.5ex}1.4 \rule[-1ex]{0pt}{0pt}& 3.5 \\
\hline
\rule{0pt}{2.5ex}1.5 \rule[-1ex]{0pt}{0pt}& 4.0 \\
\hline
\end{array}

  1. Construct an appropriate graph using the data provided, and from this, determine the threshold frequency of potassium.   (3 marks)

  1. Using the particle model of light, explain the features shown in the experimental results.   (3 marks)

Show Answers Only

a.    
         
          

b.   Features shown in experiment results:

  • Light is made of discrete energy packets called photons.
  • A photon’s energy is directly proportional to its frequency: \(E = hf\).
  • When photons hit a metal surface, they transfer their energy to electrons.
  • If the photon energy is greater than the metal’s work function \((\phi)\), the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation  \(KE = hf-\phi\).
  • Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
  • A larger stopping voltage is therefore needed to halt these more energetic electrons.
Show Worked Solution

a.    
         
       

b.   Features shown in experiment results:

  • Light is made of discrete energy packets called photons.
  • A photon’s energy is directly proportional to its frequency: \(E = hf\).
  • When photons hit a metal surface, they transfer their energy to electrons.
  • If the photon energy is greater than the metal’s work function \((\phi)\), the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation  \(KE = hf-\phi\).
  • Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
  • A larger stopping voltage is therefore needed to halt these more energetic electrons.

PHYSICS, M7 2024 HSC 32

Many scientists have performed experiments to explore the interaction of light and matter.

Analyse how evidence from at least THREE such experiments has contributed to our understanding of physics.   (8 marks)

Show Answers Only

Students could include any of the following experiments:

  • Black body radiation experiments (M7 Quantum Nature of Light)
  • Photoelectric experiments (M7 Quantum Nature of Light)
  • Spectroscopy experiments (M8 Origins of Elements)
  • Polarisation experiments (M7 Wave Nature of Light)
  • Interference and diffraction (M7 Wave Nature of Light)
  • Cosmic gamma rays (M7 Special Relativity and/or M8 Deep Inside the Atom and standard model).

Young’s Double-Slit Experiment:

  • Young’s 1801 double slit experiment aimed to determine light’s wave-particle nature.
  • He passed coherent light through two slits and observed the pattern on a screen.
  • Instead of Newton’s predicted two bright bands, Young observed alternating bright and dark bands.
  • This interference pattern occurred due to light diffraction and interference, which  re wave properties.
  • The experiment provided strong evidence for light behaving as a wave at macroscopic scales. 

Planck and the Blackbody Radiation Crisis:

  • Late 19th century scientists studied the relationship between black body radiation’s wavelength and intensity.
  • Experimental observations showed intensity peaked at a specific wavelength, contradicting classical physics predictions.
  • Classical physics led to the “ultraviolet catastrophe,” which violated energy conservation.
  • Planck’s thought experiment resolved this by proposing energy was transferred in discrete packets (quanta) where  \(E=hf\).
  • This revolutionary idea marked a shift from classical physics to quantum theory. 

Einstein and the Photoelectric Effect:

  • In 1905, Einstein built upon Plank’s idea of quantised energy to propose that light was made up of quantised photons where \(E=hf\).
  • Einstein proposition explained why electrons are ejected from metal surfaces only when light exceeds a minimum frequency.
  • Previous to Einstein’s explanation of the photoelectric effect a high intensity of light corresponds to a high energy.
  • Einstein proposed that the KE of the emitted electrons was proportion to the frequency of the light rather than the intensity of the light. 
  • This development in the understanding of the interaction of light and matter at the atomic level shifted our understanding of light to a wave-particle duality model.

Cosmic Ray Experiments and the development of the Standard Model:

  • In 1912, Victor Hess discovered cosmic rays through high-altitude balloon experiments, finding that radiation increased with altitude rather than decreased as expected.
  • The study of cosmic rays led to the unexpected discovery of new particles, including the positron and muon, which couldn’t be explained by the known models of matter.
  • These discoveries from cosmic rays helped inspire the development of modern particle accelerators and contributed to the formulation of the quark model in the 1960s.
  • Eventually further studies on these newly discovered particles led to the development of the Standard Model of particle physics, which organises all known elementary particles and their interactions.

Show Worked Solution

Students could include any of the following experiments:

  • Black body radiation experiments (M7 Quantum Nature of Light)
  • Photoelectric experiments (M7 Quantum Nature of Light)
  • Spectroscopy experiments (M8 Origins of Elements)
  • Polarisation experiments (M7 Wave Nature of Light)
  • Interference and diffraction (M7 Wave Nature of Light)
  • Cosmic gamma rays (M7 Special Relativity and/or M8 Deep Inside the Atom and standard model).

Young’s Double-Slit Experiment:

  • Young’s 1801 double slit experiment aimed to determine light’s wave-particle nature.
  • He passed coherent light through two slits and observed the pattern on a screen.
  • Instead of Newton’s predicted two bright bands, Young observed alternating bright and dark bands.
  • This interference pattern occurred due to light diffraction and interference, which  re wave properties.
  • The experiment provided strong evidence for light behaving as a wave at macroscopic scales. 

Planck and the Blackbody Radiation Crisis:

  • Late 19th century scientists studied the relationship between black body radiation’s wavelength and intensity.
  • Experimental observations showed intensity peaked at a specific wavelength, contradicting classical physics predictions.
  • Classical physics led to the “ultraviolet catastrophe,” which violated energy conservation.
  • Planck’s thought experiment resolved this by proposing energy was transferred in discrete packets (quanta) where  \(E=hf\).
  • This revolutionary idea marked a shift from classical physics to quantum theory. 

Einstein and the Photoelectric Effect:

  • In 1905, Einstein built upon Plank’s idea of quantised energy to propose that light was made up of quantised photons where \(E=hf\).
  • Einstein proposition explained why electrons are ejected from metal surfaces only when light exceeds a minimum frequency.
  • Previous to Einstein’s explanation of the photoelectric effect a high intensity of light corresponds to a high energy.
  • Einstein proposed that the KE of the emitted electrons was proportion to the frequency of the light rather than the intensity of the light. 
  • This development in the understanding of the interaction of light and matter at the atomic level shifted our understanding of light to a wave-particle duality model.

Cosmic Ray Experiments and the development of the Standard Model:

  • In 1912, Victor Hess discovered cosmic rays through high-altitude balloon experiments, finding that radiation increased with altitude rather than decreased as expected.
  • The study of cosmic rays led to the unexpected discovery of new particles, including the positron and muon, which couldn’t be explained by the known models of matter.
  • These discoveries from cosmic rays helped inspire the development of modern particle accelerators and contributed to the formulation of the quark model in the 1960s.
  • Eventually further studies on these newly discovered particles led to the development of the Standard Model of particle physics, which organises all known elementary particles and their interactions.
♦ Mean mark 50%.

PHYSICS, M7 2024 HSC 16 MC

The graph shows the relationship between the maximum kinetic energy of emitted photoelectrons and the incident photon energy for four different metal surfaces.
 

Light of frequency \(7 \times 10^{14}\ \text{Hz}\) is incident on the metals.

From which metals are photoelectrons emitted?

  1. \(\ce{K}\), \(\ce{Li}\) only
  2. \(\ce{Mg}\), \(\ce{Ag}\) only
  3. All of the metals
  4. None of the metals
Show Answers Only

\(A\)

Show Worked Solution
  • The photon energy for a light frequency of \(7 \times 10^{14}\ \text{Hz}\) is:
  •    \(E=hf = 6.626 \times 10^{-34} \times 7 \times 10^{14} = 4.64 \times 10^{-19}\ \text{J}\)
  • The energy of the light frequency in electron volts\(=\dfrac{4.64 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.9\ \text{eV}\).
  • Photoelectrons will only be emitted if the energy of the incident photons is enough to overcome the work function of the metals. On the graph above, the work function of each metal is the same value as the x-intercepts of their respective graphs.
  • Therefore photoelectrons will be emitted from potassium and lithium as the energy of the incident photons \((2.9\ \text{eV})\) is greater than the work functions of these metals.

\(\Rightarrow A\)

♦ Mean mark 47%.

PHYSICS, M7 2024 HSC 6 MC

The photoelectric effect is mathematically modelled by the following relationship:

\(K_{\max }=h f-\phi\)

In this model, the symbol \(\phi\) represents the amount of energy

  1. supplied by a photon to an electron.
  2. retained by an electron after being hit.
  3. required to release an electron from a material.
  4. left over after a collision of a photon with an electron.
Show Answers Only

\(C\)

Show Worked Solution
  • \(\phi\) is the symbol for the work function of the metal. 
  • The work function of a metal is defined as the minimum energy require for an electron to break free from the metal.

\(\Rightarrow C\)

PHYSICS, M7 2019 VCE 16

Students are studying the photoelectric effect using the apparatus shown in Figure 1.
 

     

Figure 2 shows the results the students obtained for the maximum kinetic energy \((E_{\text{k max }})\) of the emitted photoelectrons versus the frequency of the incoming light.
 

  1. Using only data from the graph, determine the values the students would have obtained for
    1. Planck's constant, \(h\). Include a unit in your answer.  (2 marks)
    1. the maximum wavelength of light that would cause the emission of photoelectrons.   (1 mark)
    1. the work function of the metal of the photocell.   (1 mark)
  1. The work function for the original metal used in the photocell is \(\phi\).
    On Figure 3, draw the line that would be obtained if a different metal, with a work function of \(\dfrac{1}{2} \phi\), were used in the photocell. The original graph is shown as a dashed line.   (2 marks)
     

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a.i.  \(5.4 \times 10^{-15}\ \text{eV s}\)

   ii.  \(811\ \text{nm}\)

  iii.  \(1.9\ \text{eV}\).

b.  

Show Worked Solution

a.i.  Planck’s constant \((h)\):

  • Equal to the gradient of the line when \(E_{\text{k max}}\) is graphed against frequency.

\(\therefore h=\dfrac{\text{rise}}{\text{run}}=\dfrac{1.25-0}{6 \times 10^{14}-3.7\times 10^{-14}}=5.4 \times 10^{-15}\ \text{eV s}\)
 

a.ii.  Max wavelength = minimum frequency of emitted photoelectron.

\(\lambda=\dfrac{c}{f}=\dfrac{3 \times 10^8}{3.7 \times 10^{14}}=811\ \text{nm}\)
 

♦ Mean mark (a)(ii) 44%.

a.iii.  

   

  • The work function is the y-intercept of the graph, so by extending the graph as shown above, the work function is \(1.9\ \text{eV}\).
     

b.   Constructing the new graph:

  • The new \(y\)-intercept for the graph will be \(-0.95\ \text{eV}\)
  • The gradient of the graph will remain the same (Planck’s constant)
     

PHYSICS, M7 2019 VCE 16 MC

Students are conducting a photoelectric effect experiment. They shine light of known frequency onto a metal and measure the maximum kinetic energy of the emitted photoelectron.

The students increase the intensity of the incident light.

The effect of this increase would most likely be

  1. lower maximum kinetic energy of the emitted photoelectrons.
  2. higher maximum kinetic energy of the emitted photoelectrons.
  3. fewer emitted photoelectrons but of higher maximum kinetic energy.
  4. more emitted photoelectrons but of the same maximum kinetic energy.
Show Answers Only

\(D\)

Show Worked Solution
  • The energy of a photon (which is transferred to the photoelectron) is independent of the intensity of the light.
  • Increasing the intensity of light will not change the maximum kinetic energy of the photoelectrons.
  • However, a higher intensity of light corresponds to more photons and thus more emitted photoelectrons.

\(\Rightarrow D\)

PHYSICS, M7 2020 VCE 15

The metal surface in a photoelectric cell is exposed to light of a single frequency and intensity in the apparatus shown in Diagram A.

The voltage of the battery can be varied in value and reversed in direction.
  

  1. A graph of photocurrent versus voltage for one particular experiment is shown in Diagram B.
  2. On Diagram B, draw the trace that would result for another experiment using light of the same frequency but with triple the intensity.  (2 marks)
     

  1. What is a name given to the point labelled \(\text{A}\) on Diagram B?   (1 mark)

  2. Why does the photocurrent fall to zero at the point labelled \(\text{A}\) on Diagram B?   (1 mark)

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a.  The photocurrent will also be tripled.

b.    The stopping voltage.

c.    Zero photocurrent at \(\text{A}\):

  • The energy of the stopping voltage is equal to the most energetic photoelectrons. 
  • Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.

Show Worked Solution

a.    The photocurrent will also be tripled.

b.    The stopping voltage.
 

c.    Zero photocurrent at \(\text{A}\):

  • The energy of the stopping voltage is equal to the most energetic photoelectrons. 
  • Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.
♦♦ Mean mark (c) 32%.
COMMENT: The work function of the metal had no relevance to this question.

PHYSICS, M7 2020 VCE 16 MC

The diagram below shows a plot of maximum kinetic energy, \(E_{\text{k max}}\),  versus frequency, \(f\), for various metals capable of emitting photoelectrons.
 

 

Which one of the following correctly ranks these metals in terms of their work function, from highest to lowest in numerical value?

  1. sodium, potassium, lithium, nickel
  2. nickel, potassium, sodium, lithium
  3. potassium, nickel, lithium, sodium
  4. lithium, sodium, potassium, nickel
Show Answers Only

\(D\)

Show Worked Solution
  • \(K_{\text{max}}=hf-\Phi \ \ \Rightarrow\ \  \Phi=hf-K_{\text{max}}\)
  • The threshold frequency occurs when \(K_{\text{max}}=0.\)
  • \(\Phi=hf_{\text{thresh}}\), hence  \(\Phi \propto f_{\text{thresh}}\).
  • The greater the threshold frequency, the greater the work function. Therefore, lithium has the greatest work function.

\(\Rightarrow D\)

PHYSICS, M7 2021 VCE 15

A photoelectric experiment is carried out by students. They measure the threshold frequency of light required for photoemission to be 6.5 × 10\(^{14}\) Hz and the work function to be 3.2 × 10\(^{-19}\) J.

Using the students' measurements, what value would they calculate for Planck's constant? Outline your reasoning and show all your working. Give your answer in joule-seconds.   (3 marks)

Show Answers Only

\(h=4.9 \times 10^{-34}\ \text{J s}\)

Show Worked Solution

\(f=6.5 \times 10^{14}\ \text{Hz},\ \ E=3.2 \times 10^{-19}\ \text{J}\)

\(E=hf\ \ \Rightarrow\ \ h=\dfrac{E}{f}\)

\(\therefore h=\dfrac{3.2 \times 10^{-19}}{6.5 \times 10^{14}} = 4.92 \times 10^{-34}\ \text{J s}\)

PHYSICS, M7 2022 VCE 14*

Sam undertakes a photoelectric effect experiment using the apparatus shown in Figure 1. She uses a green filter.
 

   

Sam produces a graph of photocurrent, \(I\), in milliamperes, versus voltage, \(V\), in volts, as shown in Figure 2.
 

   

  1. Identify what point \(\text{P}\) represents on the graph in Figure 2.   (1 mark)
  1. Sam then significantly increases the intensity of the light.
  2. Sketch the resulting graph on Figure 3. The dashed line in Figure 14 represents the original data.   (2 marks)
     

   

  1. Sam replaces the green filter with a violet filter, keeping the light source at the increased intensity.
  2. Sketch the resulting graph on Figure 4. The dashed line in Figure 4 represents the original data.   (2 marks)
     

Show Answers Only

a.    Stopping voltage        

b.   

c.   

Show Worked Solution

a.    Stopping voltage
 

b.    Diagram will exhibit the following points:

  • Increasing the intensity of light will increase the photocurrent due to more photoelectrons being ejected from the metal.
  • However, it will not change the energy of the photoelectrons, hence the stopping voltage remains the same.
     

c.    Diagram will exhibit the following points:

  • Changing the colour from green to violet will increase the frequency of the light.
  • As \(E=hf\), the energy of the photoelectrons will increase, hence there needs to be a greater stopping voltage.
     

PHYSICS, M7 2021 VCE 16 MC

The diagram below shows a circuit that is used to study the photoelectric effect.
 

Which one of the following is essential to the measurement of the maximum kinetic energy of the emitted photoelectrons?

  1. the level of brightness of the light source
  2. the wavelengths that pass through the filter
  3. the reading on the voltmeter when the current is at a minimum value
  4. the reading on the ammeter when the voltage is at a maximum value
Show Answers Only

\(C\)

Show Worked Solution
  • The maximum kinetic energy of the photoelectrons is determined by the stopping voltage. The stopping voltage can be determined by reading the voltmeter when the current is at zero.
  • \(B\) is incorrect because although knowing the wavelengths of the light passing through will help you determine the total energy of the photons, this is different to the maximum kinetic energy of the photoelectrons (where the work function needs to be taken into account).
♦♦ Mean mark 35%.

PHYSICS, M7 2022 VCE 15 MC

Which one of the following best provides evidence of light behaving as a particle?

  1. photoelectric effect
  2. white light passing through a prism
  3. diffraction of light through a single slit
  4. interference of light passing through a double slit
Show Answers Only

\(A\)

Show Worked Solution
  • During the photoelectric effect, light transfers energy in small discrete ‘quanta’ which demonstrates light behaving as a particle.

\(\Rightarrow A\)

PHYSICS, M7 2023 HSC 15 MC

What evidence resulting from investigations into the photoelectric effect is consistent with the model of light subsequently proposed by Einstein?

  1. Photoelectrons were only ejected from a metal if the light was less than a specific wavelength.
  2. Increasing the intensity of light on a metal increased the maximum kinetic energy of the photoelectrons.
  3. If photons had sufficient energy to eject photoelectrons from a metal, the maximum kinetic energy was independent of the type of metal used.
  4. The probability of photoelectrons being emitted from a metal was proportional to the duration of exposure to light for any given wavelength used.
Show Answers Only

\(A\)

Show Worked Solution
  • A photoelectron will be ejected from a metal if the energy of the photoelectron transferred from the photon is greater than the work function of the metal surface where
  •    \(K_{\text{max}}=hf-\phi\ \ \Rightarrow \ \ K_{\text{max}}=\dfrac{hc}{\lambda}-\phi\)
  • By decreasing the wavelength of the light, the maximum kinetic energy of the photoelectron increases, thus it has enough energy to overcome the work function of the metal and be ejected from the surface of the metal.
  • Increasing the intensity of the light increases the number of photons but has no effect on the maximum kinetic energy of the photoelectrons.

\(\Rightarrow A\)

♦ Mean mark 46%.

PHYSICS, M7 EQ-Bank 28

In an experiment to investigate the photoelectric effect, a group of students used a piece of equipment containing a metal cathode inside a glass tube. The students were able to accurately measure both the current produced and the maximum energy of electrons in response to light hitting the cathode.

Explain how the choice of independent variable would give rise to different results. Sketch graphs to illustrate your answer.  (7 marks)

Show Answers Only

Variables: the frequency of incident light (independent), and the maximum kinetic energy of ejected electrons (dependent).

  • Students would observe that below a certain frequency, no photoelectrons would be ejected. Photons with frequency less than the threshold frequency do not have enough energy to eject an electron.
  • Above this frequency, the students would observe that as the frequency increases, the kinetic energy of ejected electrons would increase linearly. This is because a specific amount of the photon’s energy is required to eject an electron, and any photon energy remaining is transferred to the electrons as kinetic energy, consistent with  `K_(max)=hf-Phi`.
     


 

Variables: intensity of incident light (independent) and the resultant photocurrent (dependent).

  • The frequency of light would be controlled and would be above the threshold frequency.
  • They would observe as the intensity of light increases the current produced would increase linearly.
  • This is because an increasing intensity of light increases the number of photons. This increases the rate at which photons strike the metal surface which increases the rate of photoelectron emission which in turn increases the photocurrent.
     

Show Worked Solution

Variables: the frequency of incident light (independent), and the maximum kinetic energy of ejected electrons (dependent).

  • Students would observe that below a certain frequency, no photoelectrons would be ejected. Photons with frequency less than the threshold frequency do not have enough energy to eject an electron.
  • Above this frequency, the students would observe that as the frequency increases, the kinetic energy of ejected electrons would increase linearly. This is because a specific amount of the photon’s energy is required to eject an electron, and any photon energy remaining is transferred to the electrons as kinetic energy, consistent with  `K_(max)=hf-Phi`.
     


 

Variables: intensity of incident light (independent) and the resultant photocurrent (dependent).

  • The frequency of light would be controlled and would be above the threshold frequency.
  • They would observe as the intensity of light increases the current produced would increase linearly.
  • This is because an increasing intensity of light increases the number of photons. This increases the rate at which photons strike the metal surface which increases the rate of photoelectron emission which in turn increases the photocurrent.
     

PHYSICS, M7 EQ-Bank 17 MC

The graph shows the maximum kinetic energy `(K)` with which photoelectrons are emitted as a function of frequency `(f)` for two different metals `X` and `Y`.
 

The metals are illuminated with light of wavelength 450 nm.

What would be the effect of doubling the intensity of this light without changing the wavelength?

  1. For metal `X`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  2. For metal `X`, the number of photoelectrons emitted would increase but the maximum kinetic energy remains unchanged.
  3. For metals `X` and `Y`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  4. For metals `X` and `Y`, the number of photoelectrons emitted would increase but the maximum kinetic energy remains unchanged.
Show Answers Only

`B`

Show Worked Solution
  • Calculating the frequency of incident light:
  •    `v=f lambda`
  •   `f=(f)/(lambda)=(c)/(lambda)=(3 xx10^8)/(4.5 xx10^(-7))=6.67 xx10^(14)  text{Hz}` 
  • This is greater than the work function of metal `X`, but less than the work function of metal `Y`. So, when the light is incident upon both metals, photoelectron emission will occur for `X` but not `Y`.
  • Increasing the intensity of the light increases the rate at which photons strike the surface of metal `X`. This increases the rate of photoelectron emission.
  • The only way to increase the maximum kinetic energy is to increase the frequency of incident light. So, the maximum kinetic energy of photoelectrons remains unchanged. 

`=>B`

PHYSICS, M8 EQ-Bank 27

Explain how the analysis of quantitative observations contributed to the development of the concept that certain matter and energy are quantised.   (9 marks)

Show Answers Only

Experiments such as Millikan’s oil drop experiment and others testing the photoelectric effect have demonstrated that certain quantities of matter and energy are quantised which means they are multiples of some fundamental value.

Millikan’s Oil Drop Experiment

  • Millikan’s oil drop experiment was able to show that charge is quantised. 
  • Millikan levitated oil drops in an electric field by balancing the electric and gravitational forces on them. This allowed him to find the electric force acting on each oil drop, and using the mass of the oil drop he found its charge.
  • Analysing his results, he found that the charge on every oil drop was an integer multiple of `1.602 xx10^(-19) C`. This was determined to be the fundamental charge on an electron.
  • Further, with Thompson’s later discovery of the charge to mass ratio of an electron, its mass could be determined. 

Photoelectric Effect

  • Photoelectric effect experiments showed the quantum properties of light which seemingly contradicted the view of light as a wave.
  • It was found that there was a minimum frequency (energy) of light that would cause photoemission when it was incident upon a metal plate, and no photoemission occurred with light lower than this frequency, regardless of intensity.
  • As one photon would strike one electron on the metal surface, the electron would receive a discrete amount of energy from that photon determined by its frequency `E=hf`. If a photon didn’t have enough energy, an electron couldn’t be removed.
  • This experimental evidence changed the conceptual understanding of energy within physics and provided a basis for the quantisation of the energy of light. 

Other quantitative experiments that could be explored include:

  • Bohr’s analysis of emission spectra to demonstrate the existence of quantised energy levels in atoms.
  • Cathode ray experiments showing the particle nature of electrons.
  • Blackbody radiation experiments. 
Show Worked Solution

Experiments such as Millikan’s oil drop experiment and others testing the photoelectric effect have demonstrated that certain quantities of matter and energy are quantised which means they are multiples of some fundamental value.

Millikan’s Oil Drop Experiment

  • Millikan’s oil drop experiment was able to show that charge is quantised. 
  • Millikan levitated oil drops in an electric field by balancing the electric and gravitational forces on them. This allowed him to find the electric force acting on each oil drop, and using the mass of the oil drop he found its charge.
  • Analysing his results, he found that the charge on every oil drop was an integer multiple of `1.602 xx10^(-19) C`. This was determined to be the fundamental charge on an electron.
  • Further, with Thompson’s later discovery of the charge to mass ratio of an electron, its mass could be determined. 

Photoelectric Effect

  • Photoelectric effect experiments showed the quantum properties of light which seemingly contradicted the view of light as a wave.
  • It was found that there was a minimum frequency (energy) of light that would cause photoemission when it was incident upon a metal plate, and no photoemission occurred with light lower than this frequency, regardless of intensity.
  • As one photon would strike one electron on the metal surface, the electron would receive a discrete amount of energy from that photon determined by its frequency `E=hf`. If a photon didn’t have enough energy, an electron couldn’t be removed.
  • This experimental evidence changed the conceptual understanding of energy within physics and provided a basis for the quantisation of the energy of light. 

Other quantitative experiments that could be explored include:

  • Bohr’s analysis of emission spectra to demonstrate the existence of quantised energy levels in atoms.
  • Cathode ray experiments showing the particle nature of electrons.
  • Blackbody radiation experiments. 

PHYSICS, M7 2016 HSC 13 MC

When light of a specific frequency strikes a metal surface, photoelectrons are emitted.

If the light intensity is increased but the frequency remains the same, which row of the table is correct?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex}\text{} & \text{} \\
\text{}\rule[-0.5ex]{0pt}{0pt}& \text{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{1.5ex}\textit{Number of photoelectrons} & \textit{Maximum kinetic energy} \\
\textit{emitted}\rule[-0.5ex]{0pt}{0pt}& \textit{of the photoelectrons} \\
\hline
\rule{0pt}{2.5ex}\text{Remains the same}\rule[-1ex]{0pt}{0pt}&\text{Remains the same}\\
\hline
\rule{0pt}{2.5ex}\text{Remains the same}\rule[-1ex]{0pt}{0pt}& \text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Remains the same} \\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Increases} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution
  • An increase in light intensity increases the number of photons striking the metal surface.
  • This increases the number of photoelectrons emitted.
  • Since no change in frequency, the energy of photons striking metal surface remains constant.
  • Maximum kinetic energy of emitted photoelectrons remains the same.

\(\Rightarrow C\)

PHYSICS, M7 2016 HSC 27b

Explain how the result of ONE investigation of the photoelectric effect changed the scientific understanding of the nature of light.   (4 marks)

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  • One investigation of the photoelectric effect showed that the kinetic energy of emitted photoelectrons increased when the frequency of incident light increased, but did not increase with increased light intensity.
  • This could only be explained by a particle, or photon model of light.
  • The photon model considers light to consist of discrete packets of energy, where the energy of a photon is proportional to its frequency.
  • Thus, higher frequency light → greater photon energy → greater kinetic energy of emitted photoelectrons.
  • The scientific understanding of light was changed from a wave model where light existed only as transverse waves of different wavelengths to a wave-particle duality.
Show Worked Solution
  • One investigation of the photoelectric effect showed that the kinetic energy of emitted photoelectrons increased when the frequency of incident light increased, but did not increase with increased light intensity.
  • This could only be explained by a particle, or photon model of light.
  • The photon model considers light to consist of discrete packets of energy, where the energy of a photon is proportional to its frequency.
  • Thus, higher frequency light → greater photon energy → greater kinetic energy of emitted photoelectrons.
  • The scientific understanding of light was changed from a wave model where light existed only as transverse waves of different wavelengths to a wave-particle duality.

♦ Mean mark 40%.

PHYSICS, M7 2017 HSC 21

A laser emits light of wavelength 550 nm.

  1. Calculate the frequency of this light.   (2 marks)
  1. The electrons in a specific metal must absorb a minimum of  `5 × 10^(-19)\ text{J}`  in order to be ejected from its surface.
  2. Explain why electrons will not be ejected from this metal when photons of wavelength 550 nm strike its surface. Support your answer with relevant calculations.   (3 marks)
Show Answers Only

a.   `f=5.45 xx10^(14)  text{Hz}`

b.    `E` `=3.63 xx10^(-19)  text{J}`
    `<5 xx10^(-19)  text{J}`

 
`text{So, no electrons are ejected.}`

Show Worked Solution
a. `v` `=flambda`
  `f` `=(v)/(lambda)=(3xx10^(8))/(550 xx10^(-9))=(3xx10^(8))/(550 xx10^(-9))`

 

b. `E` `=hf`
    `=6.62 xx10^(-34)xx5.46 xx10^(14)`
    `=3.63 xx10^(-19)  text{J}`
     
  •  The work function of the metal sample is  `5 xx10^(-19)\ text{J}`.
  • Since the incident photon energy of  `3.63 xx10^(-19)  text{J}`  is less than  `5 xx10^(-19)\ text{J}`, they are unable to eject electrons from the metal.

PHYSICS, M7 2018 HSC 17 MC

The graph shows the maximum kinetic energy of electrons ejected from different metals as a function of the frequency of the incident light.
 

What can be deduced from this graph?

  1. The maximum kinetic energy of ejected electrons is proportional to the number of photons incident on the metal surface.
  2. More photons are required to cause an electron to be ejected from zinc than from potassium.
  3. Any photon that can eject an electron from the surface of zinc must also be able to cause an electron to be ejected from potassium.
  4. For any given frequency that causes electrons to be ejected from all three metals, the number of electrons ejected is always greatest for potassium.
Show Answers Only

`C`

Show Worked Solution
  • The graph shows that zinc has a greater work function than potassium.
  • So, any photon with sufficient energy to eject a photon from the surface of zinc will also have sufficient energy to overcome the work function of potassium and eject a photon from its surface.

`=>C`

PHYSICS, M7 2013 HSC 20 MC

The graph shows the maximum kinetic energy `E` with which photoelectrons are emitted as a function of frequency `f` for two different metals `X` and `Y`.
 


 

The metals are illuminated with light of wavelength 450 nm.

What would be the effect of doubling the intensity of this light without changing the wavelength?

  1. For metal `X`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  2. For metal `X`, the number of photoelectrons emitted would increase but the maximum kinetic energy would remain unchanged.
  3. For both metals `X` and `Y`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  4. For both metals `X` and `Y`, the number of photoelectrons emitted would increase but the maximum kinetic energy would remain unchanged.
Show Answers Only

 

Show Worked Solution

PHYSICS M7 2022 HSC 26

Light of frequency  `7.5 xx10^(14) \ text{Hz}`  is incident on a calcium metal sheet which has a work function of  `2.9 \ text{eV}`. Photoelectrons are emitted.

The metal is in a uniform electric field of `5.2 \ text{NC}^{-1}`, perpendicular to the surface of the metal, as shown.
 


  

  1. Show that the maximum kinetic energy of an emitted photoelectron is  `3.2 xx10^(-20) \ text{J}`.  (3 marks)
  1. Calculate the maximum distance, `d`, an emitted photoelectron can travel from the surface of the metal.  (3 marks)
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  1. Proof (See Worked Solution)
  2. `0.038  text{m.}`
Show Worked Solution
a.    `K_(max)` `=hf-Phi`
    `=6.626 xx10^(-34)xx7.5 xx10^(14)-2.9xx 1.602 xx10^(-19)`
    `=3.2 xx10^(-20)  text{J}`

 

b.   Maximum Distance:

  • The electric field will do `3.2 xx10^(-20)\ text{J}` of work to stop the electron.
`W` `=qEd`     
`3.2 xx10^(-20)` `=1.602 xx10^(-19) xx 5.2 xx d_max`  
`:.d_max` `=(3.2 xx10^(-20))/(1.602 xx10^(-19)xx 5.2)`  
  `=0.038  text{m.}`  

♦ Mean mark (b) 51%.

PHYSICS, M7 2022 HSC 14 MC

Line `X` shows the results of an experiment carried out to investigate the photoelectric effect.
 


 

What change to this experiment would produce the results shown by line `Y` ?

  1. Increasing the frequency of the radiation
  2. Using a metal that has a greater work function
  3. Decreasing the intensity of the incident radiation
  4. Decreasing the maximum energy of photoelectrons
Show Answers Only

`B`

Show Worked Solution
  • The work function of the metals is given by the respective y-intercepts of lines `X` and `Y`. 
  • As line `Y` has a y-intercept with greater magnitude, it can be produced by using a metal with a greater work function.

`=>B`

PHYSICS, M7 2019 HSC 23

A student investigated the photoelectric effect. The frequency of light incident on a metal surface was varied and the corresponding maximum kinetic energy of the photoelectrons was measured.

The following results were obtained.
 

\begin{array}{|l|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\textit{Frequency}\ \left(\times 10^{14} Hz\right) \rule[-1ex]{0pt}{0pt}& 11.2 & 13.5 & 15.2 & 18.6 & 20.0 \\
\hline \rule{0pt}{2.5ex}\textit{Maximum kinetic energy}\ (eV) \rule[-1ex]{0pt}{0pt}& 0.6 & 1.3 & 2.3 & 3.3 & 4.2 \\
\hline
\end{array}

Plot the results on the axes below and hence determine the work function of the metal in electron volts.   (3 marks)
 

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Work function = 4 eV

Show Worked Solution

The metal has a work function of 4 eV (this is the negative of the \(y\)-intercept).

PHYSICS, M7 2019 HSC 13 MC

A laser has a power output of 30 mW and emits light with a wavelength of 650 nm.

How many photons does this laser emit per second?

  1. `4.6 × 10^(14)`
  2. `9.8 × 10^(16)`
  3. `3.1 × 10^(19)`
  4. `9.3 × 10^(21)`
Show Answers Only

`B`

Show Worked Solution
`E` `=hf`  
  `=(hc)/(lambda)`  
  `=((6.626 xx10^(-34))(3 xx10^(8)))/(6.5 xx10^(-7))`  
  `=3.1 xx10^(-19)\ \text{J}`  

 
`text{Photons}=(P)/(E)=(30 xx10^(-3))/(3.1 xx10^(-19))=9.8 xx10^(16)`

`=>B`


♦ Mean mark 43%.

PHYSICS, M7 2020 HSC 13 MC

The graph shows the relationship between the frequency of light used to irradiate two different metals, and the maximum kinetic energy of photoelectrons emitted.
 


 

Suppose that light having a frequency of `8 × 10^(14)` Hz is used to irradiate both metals.

Compared to the photoelectrons emitted from metal `X`, photoelectrons emitted from metal `Y` will

  1. have a lower maximum velocity.
  2. have a higher maximum velocity.
  3. take a longer time to gain sufficient energy to be ejected.
  4. take a shorter time to gain sufficient energy to be ejected.
Show Answers Only

`A`

Show Worked Solution
  • The graph shows photoelectrons emitted from metal ` Y` have a lower maximum kinetic energy. They therefore have a lower maximum velocity.
  • The time taken for photoelectrons to gain sufficient energy to be ejected is the same for both metals.

`=>A`


♦ Mean mark 49%.

PHYSICS, M7 2021 HSC 20 MC

A metal cylinder is located in a uniform magnetic field. The work function of the metal is `phi`.

Photons having an energy of 2`phi` strike the side of the cylinder, liberating photoelectrons which travel perpendicular to the magnetic field in a circular path. The maximum radius of the path is `r`.
 

If the photon energy is doubled, what will the maximum radius of the path become?

  1. `2r`
  2. `3r`
  3. `sqrt2r`
  4. `sqrt3r`
Show Answers Only

`D`

Show Worked Solution
`K_max` `=(1)/(2)mv_(max)^2`  
`v_(max)` `=sqrt((2K_(max))/(m))\ \ `… (1)  

  
Substitute (1) into  `r=(mv)/(qB)`:

    `r=(m)/(qB)sqrt((2K_(max))/(m))`

    `r prop sqrt(K_(max))`
 

  • When the photon energy is `2phi, K_max=phi.`
  • When the photon energy is doubled to `4phi, K_max=3phi.`
  • ∴ Since  `r prop sqrt(K_(max)) ` the radius increases by a factor of `sqrt(3).`

`=>D`


♦ Mean mark 24%.