2019 A/L Physics Past Paper · Question 32 (English Medium)

Photoelectric effect: the longest wavelength

Part of an orbiting satellite is coated with a metal that has a work function of 5 eV5\ \text{eV}. The Planck constant is 4.1×10−15 eV s4.1 \times 10^{-15}\ \text{eV s} and the speed of light is 3×108 m s−13 \times 10^8\ \text{m s}^{-1}. What could be the longest wavelength of incident sunlight that can eject an electron from the metal coating?

  1. (1) 12.3 nm
  2. (2) 246 nm
  3. (3) 683 nm
  4. (4) 800 nm
  5. (5) 1230 nm
Answer and explanation

Answer: (2) 246 nm

Given information

We know:

The work function φ=5\varphi = 5 eV.

h=4.1×10−15h = 4.1 \times 10^{-15} eV s.

c=3×108 m s−1c = 3 \times 10^8\ \text{m s}^{-1}.

What the question asks

We want:

The longest wavelength λmax\lambda_{max} that can free an electron.

At this wavelength the photon has just enough energy.

Relevant theory

where h is the Planck constant and f the frequency of the absorbed photon.

Light comes as photons.

One photon has the energy:

E=hf=hcλE = hf = \dfrac{hc}{\lambda}

One photon gives all its energy to one electron.

To get out, the electron needs at least the work function φ\varphi.

A longer wave gives less energy.

So the longest wave is the one with E=φE = \varphi.

The energy required to get an electron free from the metal surface is known as the work function of that metal.

Required equations

E=hf=hcλE = hf = \frac{hc}{\lambda}

hcλmax=φ⇒λmax=hcφ\frac{hc}{\lambda_{max}} = \varphi \Rightarrow \lambda_{max} = \frac{hc}{\varphi}

Applying it to this question

Units:

hh is in eV s and φ\varphi is in eV.

The eV cancels. So λ\lambda comes out in metres.

There is no need to change eV into joules.

Reasoning and calculation

λmax=4.1×10−15×3×1085\lambda_{max} = \frac{4.1 \times 10^{-15} \times 3 \times 10^8}{5}

λmax=1.23×10−65\lambda_{max} = \frac{1.23 \times 10^{-6}}{5}

246246

λmax=2.46×10−7 m\lambda_{max} = 2.46 \times 10^{-7}\ \text{m}

2.46×10−72.46 \times 10^{-7} m is 246 nm.

That is option (2).

Checks

h∗c/phih * c / phi

Does it make sense?

246 nm is ultraviolet light.

Visible light is 400 nm to 700 nm. It is too long.

So only the ultraviolet part of sunlight frees electrons from this metal.

Conclusion

λmax=246 nm\lambda_{max} = 246\ \text{nm}

λmax=hcφ=1.23×10−65=246 nm\lambda_{max} = \dfrac{hc}{\varphi} = \dfrac{1.23 \times 10^{-6}}{5} = 246\ \text{nm}

So the answer is option (2).

If you remember hc≈1240hc \approx 1240 eV nm, then:

λ (nm)=1240E (eV)\lambda\,(\text{nm}) = \dfrac{1240}{E\,(\text{eV})}

Here hchc is 12301230 eV nm.

Exam tip

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