smc-3698-60-Experiments

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.

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→ 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 2014 HSC 26

Calculate the energy of a photon of wavelength 415 nm. (2 marks)
An experiment was conducted using a photoelectric cell as shown in the diagram.

The graph plots the maximum kinetic energy of the emitted photoelectrons against radiation frequency for the aluminium surface.

The experiment is planned to be repeated using a voltage of 0.0 V.

Draw a line on the graph to show the predicted results of the planned experiment, and determine the radiation frequency which would produce photoelectrons with a maximum kinetic energy of 1.2 eV using a voltage of 0.0 V. (3 marks)

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`text(See Worked Solutions)`

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a.	`c`	`= f lambda`
		`= hf`
		`= (hc)/lambda`
		`= (6.626 xx 10^(-34) Js xx 3 xx 10^8 ms^(-1))/(415 xx 10^(-9)m)`
		`= 4.79 xx 10^(-19) J`

`text(To find intercept)`

`4.1 text(V)`	`= 4.1 xx 1.602 xx 10^(-19)\ text(J)`	`text(of energy required to be)` `text(supplied by the photon.)`
	`= 6.56 xx 10^(-19)\ text(J)`
`hf`	`= 6.56 xx 10^(-19)\ text(J)`
`f`	`= (6.56 xx 10^(-19))/(6.626 xx 10^(-34))`
	`= 9.9 xx 10^14`

`text(Gradient = same as A1)`

`text{From graph maximum KE(eV)}`	`= 1.2\ text(eV)`
`text(Frequency)`	`= 12.8\ text(Hz)`
	`= 12.8 xx 10^14\ text(Hz)`

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` ?

Increasing the frequency of the radiation
Using a metal that has a greater work function
Decreasing the intensity of the incident radiation
Decreasing the maximum energy of photoelectrons

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`B`

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→ 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.

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

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

PHYSICS, M7 2020 HSC 3 MC

What was the basis for Maxwell's prediction of the velocity of electromagnetic waves?

Experiments using magnetic fields to accelerate particles
Experiments using light and mirrors to establish the finite speed of light
Equations showing how oscillating electric and magnetic fields propagate
Equations showing how electromagnetic waves are affected by gravitational fields

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`C`

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Maxwell used equations to show interactions between electric and magnetic fields.

`=>C`

PHYSICS, M7 2021 HSC 33

Two experiments are performed with identical light sources having a wavelength of 400 nm.

In experiment `A`, the light is incident on a pair of narrow slits `5.0 xx 10^(-5)` m apart, producing a pattern on a screen located 3.0 m behind the slits.

In experiment `B`, the light is incident on different metal samples inside an evacuated tube as shown. The kinetic energy of any emitted photoelectrons can be measured.

Some results from experiment `B` are shown.

How do the results from Experiment `A` and Experiment `B` support TWO different models of light? In your answer, include a quantitative analysis of each experiment. (9 marks)

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→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

`d sin theta`	`=m lambda`
`5xx10^(-5)sin theta`	`=1xx400 xx10^(-9)`
`theta`	`=0.46^(@)`
`s`	`=3 xx tan(0.46) = 0.024` m

→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`

Show Worked Solution

→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

`d sin theta`	`=m lambda`
`5xx10^(-5)sin theta`	`=1xx400 xx10^(-9)`
`theta`	`=0.46^(@)`
`s`	`=3 xx tan(0.46) = 0.024` m

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`

♦ Mean mark 52%.