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Explain the concept of light quantization, work function, atomic number, and mass number. Calculate the work function and threshold frequency for a metal given photon energy and maximum kinetic energy

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2 parts
a(iii)

3.58×1014 Hz\text{3.58} \times 10^{14} \text{ Hz}

c(ii)

86222Rn⟶88222Ra+2×−10β_{86}^{222}\text{Rn} \longrightarrow _{88}^{222}\text{Ra} + 2 \times _{-1}^{0}\beta

Step-by-step solution

3 steps

Part a(i)

Light is quantized means that its energy is not continuous but exists in discrete packets called photons. Each photon carries a specific amount of energy.

Part a(ii)

The work function (ϕ\phi) is the minimum amount of energy required to remove an electron from the surface of a metal.

Part a(iii)

3.58×1014 Hz\text{3.58} \times 10^{14} \text{ Hz}

  1. Calculate the energy of the incident photon using the given frequency.

    E=hfE = hf E=(6.6×10−34 J s)×(5.4×1014 Hz)E = (6.6 \times 10^{-34} \text{ J s}) \times (5.4 \times 10^{14} \text{ Hz}) E=3.564×10−19 JE = 3.564 \times 10^{-19} \text{ J}

  2. Use the photoelectric equation to find the work function (ϕ\phi).

    E=ϕ+KEmaxE = \phi + KE_{max} ϕ=E−KEmax\phi = E - KE_{max} ϕ=(3.564×10−19 J)−(1.2×10−19 J)\phi = (3.564 \times 10^{-19} \text{ J}) - (1.2 \times 10^{-19} \text{ J}) ϕ=2.364×10−19 J\phi = 2.364 \times 10^{-19} \text{ J}

  3. Calculate the minimum frequency (threshold frequency, f0f_0) using the work function.

    ϕ=hf0\phi = hf_0 f0=ϕhf_0 = \frac{\phi}{h} f0=2.364×10−19 J6.6×10−34 J sf_0 = \frac{2.364 \times 10^{-19} \text{ J}}{6.6 \times 10^{-34} \text{ J s}} f0=3.58×1014 Hzf_0 = 3.58 \times 10^{14} \text{ Hz}

    The minimum frequency of radiation is 3.58×1014 Hz\boxed{\text{3.58} \times 10^{14} \text{ Hz}}.

Part b(i)

The atomic number (Z) is the number of protons found in the nucleus of an atom. It determines the element's identity.

Part b(ii)

The mass number (A) is the total number of protons and neutrons (nucleons) in the nucleus of an atom.

Part c(i)

The symbol for radon nuclide is 86222Rn\boxed{_{86}^{222}\text{Rn}}.

Part c(ii)

86222Rn⟶88222Ra+2×−10β_{86}^{222}\text{Rn} \longrightarrow _{88}^{222}\text{Ra} + 2 \times _{-1}^{0}\beta

The decay process involves the emission of two beta particles (−10β_{-1}^{0}\beta).

86222Rn⟶88222Ra+2×−10β_{86}^{222}\text{Rn} \longrightarrow _{88}^{222}\text{Ra} + 2 \times _{-1}^{0}\beta

The equation representing the decay process is 86222Rn⟶88222Ra+2×−10β\boxed{_{86}^{222}\text{Rn} \longrightarrow _{88}^{222}\text{Ra} + 2 \times _{-1}^{0}\beta}.

Part d

Three properties of x-rays are:

  • They are a form of electromagnetic radiation.
  • They travel at the speed of light in a vacuum.
  • They are not deflected by electric or magnetic fields.
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