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What is a photon? (6) An investigation was carried out to establish the relationship between the current flowing in a photocell and the frequency of the light incident on it - Leaving Cert Physics - Question 8 - 2009

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What is a photon? (6) An investigation was carried out to establish the relationship between the current flowing in a photocell and the frequency of the light incid... show full transcript

Worked Solution & Example Answer:What is a photon? (6) An investigation was carried out to establish the relationship between the current flowing in a photocell and the frequency of the light incident on it - Leaving Cert Physics - Question 8 - 2009

Step 1

What is a photon?

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Answer

A photon is defined as a packet of energy or quantum of electromagnetic radiation. It is considered to be the fundamental unit of light and other forms of electromagnetic radiation, characterized by its energy that is proportional to its frequency.

Step 2

Draw a labelled diagram of the structure of a photocell.

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Answer

A typical photocell structure consists of:

  • A cylindrical cathode (electrode) surrounded by a glass casing
  • A wire anode (electrode)
  • A vacuum region within the casing. Ensure each part is clearly labeled in the diagram for clarity.

Step 3

Using the graph, calculate the work function of the metal.

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Answer

The work function (φ) can be calculated using the equation:

φ=hfφ = hf

where:

  • h=6.6imes1034Jsh = 6.6 imes 10^{-34} J s (Planck's constant)
  • f=5.2imes1014Hzf = 5.2 imes 10^{14} Hz (frequency from the graph) Therefore,
φ=(6.6imes1034)(5.2imes1014)ightarrowφ=3.432imes1019J.φ = (6.6 imes 10^{-34})(5.2 imes 10^{14}) ightarrow φ = 3.432 imes 10^{-19} J.

Step 4

What is the maximum speed of an emitted electron when light of wavelength 550 nm is incident on the photocell?

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Answer

To find the maximum speed (v) of the emitted electron, use the equation for conservation of energy:

hf = rac{1}{2} mv^2 \\ \text{Where:} \\ - h = 6.6 \times 10^{-34} J s \\ - f = \frac{3 \times 10^{8} m/s}{550 \times 10^{-9} m} = 5.45 \times 10^{14} Hz \\ - m = 9.1 \times 10^{-31} kg \\ \text{Substituting values:}

Calculating, we find:

v=1.922×105m/s.v = 1.922 \times 10^{5} m/s.

Step 5

Explain why a current does not flow in the photocell when the frequency of light is less than 5.2 x 10^14 Hz.

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Answer

A current does not flow in the photocell when the frequency of the incident light is below the threshold frequency because the energy of the incoming photons is insufficient to overcome the work function of the metal. Hence, no electrons are emitted.

Step 6

Draw a sketch of the graph obtained. How was the intensity of the light varied?

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Answer

The graph should display intensity (I) on the y-axis against distance (or intensity) on the x-axis with a linear relationship illustrating increased current with increased light intensity. The intensity of light can be varied by changing the distance of the light source to the photocell or altering the light source power.

Step 7

What conclusion about the nature of light can be drawn from these investigations?

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Answer

The investigations conclude that light has a quantized nature, demonstrating that it behaves as both a wave and a particle. The phenomenon of photoelectric effect showcases that light is made up of photons, or bundles of energy, reinforcing its particle-like properties. Thus, we can say that light has a corpuscular nature, while also exhibiting wave-like behavior.

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