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Figure 1 shows an arrangement used to investigate the photoelectric effect - AQA - A-Level Physics - Question 2 - 2020 - Paper 1

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Figure 1 shows an arrangement used to investigate the photoelectric effect. A current is measured on the microammeter only when electromagnetic radiation with a fre... show full transcript

Worked Solution & Example Answer:Figure 1 shows an arrangement used to investigate the photoelectric effect - AQA - A-Level Physics - Question 2 - 2020 - Paper 1

Step 1

Explain why the frequency of the electromagnetic radiation must be greater than a certain value.

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Answer

The frequency of the electromagnetic radiation must exceed a certain threshold frequency because it is directly related to the energy of the photons. According to the equation for the energy of a photon, we have:

E=hfE = h f

where:

  • EE is the energy of the photon,
  • hh is Planck's constant, and
  • ff is the frequency of the radiation.

To remove an electron from the surface of the photoemissive material, the energy of the incoming photon must be equal to or greater than the work function Ï•\phi of the material. Therefore, the relationship can be expressed as:

hf≥ϕh f \geq \phi

If the frequency is lower than this threshold, the photons will not have sufficient energy to dislodge electrons, resulting in no current in the circuit.

Step 2

Determine the number of photoelectrons per second leaving the photoemissive surface when the current is a maximum.

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Answer

The number of photoelectrons released per second corresponds to the maximum current measured in the circuit, given by the relationship:

I=neI = n e

where:

  • II is the current,
  • nn is the number of photoelectrons released per second, and
  • ee is the elementary charge (approximately 1.6×10−191.6 \times 10^{-19} C).

At the maximum current, if, for instance, II is given as 1.9μA1.9 \mu A, then:

  1. Convert II to amperes: I=1.9×10−6AI = 1.9 \times 10^{-6} A

  2. Rearranging the equation gives: n=Ie=1.9×10−61.6×10−19≈1.19×1013 electrons per secondn = \frac{I}{e} = \frac{1.9 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 1.19 \times 10^{13} \text{ electrons per second}

Step 3

Explain why I reaches a constant value for positive values of V.

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Answer

The current II reaches a constant value for positive values of the potential difference VV because all emitted photoelectrons are effectively collected at the anode. When a positive potential is applied, it creates an electric field that accelerates the photoelectrons towards the anode. As the potential increases, more electrons are attracted, and eventually, a point is reached where all emitted electrons are captured, resulting in a saturated current. At saturation, the current no longer increases with further increases in VV, hence it remains constant.

Step 4

Explain why I decreases as the value of V' becomes more negative.

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Answer

As the potential difference V′V' becomes more negative, fewer photoelectrons can reach the anode due to the opposing electric field. The negative potential repels the electrons emitted from the photoemissive surface, reducing the number of electrons that can be collected. Consequently, this leads to a decrease in the measured current II. If the negative potential becomes sufficiently large, it may eventually prevent any emitted photoelectrons from reaching the anode, resulting in zero current.

Step 5

Discuss the effect that this change in surface has on the value of the stopping potential.

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Answer

When the investigation is repeated with a different photoemissive surface that has a smaller value of the work function Ï•\phi, the stopping potential will also change. The stopping potential is defined by the amount of potential needed to stop the most energetic photoelectrons from reaching the anode. Since the work function is lower, it implies that the energy of the emitted photoelectrons is higher for the same incident light frequency. Therefore, the required stopping potential decreases, as there are more energetic electrons that can be stopped with a lesser potential. As a result, the overall stopping potential will diminish when utilizing a material with a smaller work function.

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