A-level Physics
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    8.1 Photons and the photoelectric effect — Practise with help

    Work each problem first. Open a hint only if you are stuck, and the full working only to check. Change nanometres to metres and electronvolts to joules before you substitute.

    Warm-up: remember

    1. A wave has a frequency of 6.0×1014 Hz6.0 \times 10^{14}\,\text{Hz}. Find its wavelength. The speed of light is 3.00×108 m s−13.00 \times 10^{8}\,\text{m}\,\text{s}^{-1}.
    2. Write an energy of 2.0 eV2.0\,\text{eV} in joules.
    3. An electron is accelerated through a potential difference of 50 V50\,\text{V}. Find the kinetic energy it gains, in joules.
    Answers
    1. λ=cf=3.00×1086.0×1014=5.0×10−7 m\lambda = \dfrac{c}{f} = \dfrac{3.00 \times 10^{8}}{6.0 \times 10^{14}} = 5.0 \times 10^{-7}\,\text{m}.
    2. E=2.0×1.60×10−19=3.2×10−19 JE = 2.0 \times 1.60 \times 10^{-19} = 3.2 \times 10^{-19}\,\text{J}.
    3. Ek=eV=1.60×10−19×50=8.0×10−18 JE_k = eV = 1.60 \times 10^{-19} \times 50 = 8.0 \times 10^{-18}\,\text{J}.

    If question 1 goes wrong, revisit the wave equation in Section 6.1. If question 2 goes wrong, remember that one electronvolt is 1.60×10−19 J1.60 \times 10^{-19}\,\text{J}. If question 3 goes wrong, the energy gained is the charge multiplied by the potential difference.

    Practise with help

    1. A photon has a frequency of 5.0×1014 Hz5.0 \times 10^{14}\,\text{Hz}. Find its energy.

    Hint

    The energy of a photon is the Planck constant multiplied by the frequency.

    Next step

    Write E=hfE = hf with h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J}\,\text{s} and f=5.0×1014 Hzf = 5.0 \times 10^{14}\,\text{Hz}.

    Full working E=hf=6.63×10−34×5.0×1014=3.3×10−19 J\begin{aligned} E &= hf \\ &= 6.63 \times 10^{-34} \times 5.0 \times 10^{14} \\ &= 3.3 \times 10^{-19}\,\text{J} \end{aligned}

    The photon energy is 3.3×10−19 J3.3 \times 10^{-19}\,\text{J}, to two significant figures.

    2. Find the energy of a photon of wavelength 400 nm, in joules and in electronvolts.

    Hint

    Use E=hcλE = \dfrac{hc}{\lambda}. Change the wavelength from nanometres to metres first.

    Next step

    Write λ=4.00×10−7 m\lambda = 4.00 \times 10^{-7}\,\text{m}, then divide hchc by it. Divide the answer in joules by 1.60×10−191.60 \times 10^{-19} to get electronvolts.

    Full working E=hcλ=6.63×10−34×3.00×1084.00×10−7=4.97×10−19 J=4.97×10−191.60×10−19=3.11 eV\begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{4.00 \times 10^{-7}} \\ &= 4.97 \times 10^{-19}\,\text{J} \\ &= \frac{4.97 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.11\,\text{eV} \end{aligned}

    The photon energy is 4.97×10−19 J4.97 \times 10^{-19}\,\text{J}, or 3.11 eV.

    3. A metal has a work function of 2.30 eV. Light of wavelength 300 nm falls on it. Find the maximum kinetic energy of the emitted electrons, in eV.

    Hint

    Use the photoelectric equation. Keep both energies in electronvolts.

    Next step

    Find the photon energy in eV with hcλ\dfrac{hc}{\lambda}, then subtract the work function.

    Full working E=hcλ=6.63×10−34×3.00×1083.00×10−7=6.63×10−19 J=4.14 eV\begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.00 \times 10^{-7}} \\ &= 6.63 \times 10^{-19}\,\text{J} = 4.14\,\text{eV} \end{aligned} Ek(max)=hf−ϕ=4.14−2.30=1.84 eVE_{k(\text{max})} = hf - \phi = 4.14 - 2.30 = 1.84\,\text{eV}

    The most energetic electrons leave with 1.84 eV of kinetic energy.

    4. A metal has a work function of 2.00 eV. Find its threshold wavelength.

    Hint

    At the threshold the photon energy equals the work function.

    Next step

    Change the work function to joules, then use λ0=hcϕ\lambda_0 = \dfrac{hc}{\phi}.

    Full working ϕ=2.00×1.60×10−19=3.20×10−19 Jλ0=hcϕ=6.63×10−34×3.00×1083.20×10−19=6.22×10−7 m\begin{aligned} \phi &= 2.00 \times 1.60 \times 10^{-19} = 3.20 \times 10^{-19}\,\text{J} \\ \lambda_0 &= \frac{hc}{\phi} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.20 \times 10^{-19}} \\ &= 6.22 \times 10^{-7}\,\text{m} \end{aligned}

    The threshold wavelength is 622 nm.

    5. In an experiment, the maximum kinetic energy of the photoelectrons is 2.5 eV. Find the stopping potential.

    Hint

    The stopping potential is the potential difference that brings the fastest electron to rest.

    Next step

    Use eVs=Ek(max)eV_s = E_{k(\text{max})}, and remember that an energy in electronvolts and a potential difference in volts share the same number.

    Full working eVs=Ek(max)=2.5 eVVs=2.5 VeV_s = E_{k(\text{max})} = 2.5\,\text{eV} \qquad V_s = 2.5\,\text{V}

    The stopping potential is 2.5 V.

    6. A hydrogen atom emits a photon when an electron falls from a level at −1.51 eV-1.51\,\text{eV} to a level at −3.40 eV-3.40\,\text{eV}. Find the energy of the photon in electronvolts, and its wavelength.

    Hint

    The photon energy is the difference between the two levels. Both levels are negative, so subtract carefully.

    Next step

    Find the gap in eV, change it to joules, then use E=hcλE = \dfrac{hc}{\lambda}.

    Full working E=−1.51−(−3.40)=1.89 eV=1.89×1.60×10−19=3.02×10−19 J\begin{aligned} E &= -1.51 - (-3.40) = 1.89\,\text{eV} \\ &= 1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19}\,\text{J} \end{aligned} λ=hcE=6.63×10−34×3.00×1083.02×10−19=6.58×10−7 m\begin{aligned} \lambda &= \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.02 \times 10^{-19}} \\ &= 6.58 \times 10^{-7}\,\text{m} \end{aligned}

    The photon has an energy of 1.89 eV and a wavelength of 658 nm, which is red light.

    7. Explain why increasing the intensity of light below the threshold frequency releases no photoelectrons, however bright the light is.

    Hint

    Think about what one photon can do to one electron.

    Next step

    Compare the energy of one photon with the work function. Then say what more photons of the same energy can and cannot do.

    Full working

    A photon is absorbed by one electron. If the photon energy hfhf is less than the work function ϕ\phi, the electron cannot escape, even though it gains energy for a moment. Increasing the intensity adds more photons, but each photon still carries too little energy. The electron cannot collect energy from two photons at once. No electrons are released.

    Below the threshold, the photons are too weak on their own. The work function is a one-photon barrier, not a total-energy barrier.

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