Photon Energy (DP IB Physics: SL): Revision Note

Explain how atomic spectra provide evidence for the quantisation of energy in atoms.

Answer:

Step 1: Outline the meaning of atomic spectra

• Atomic spectra show the spectrum of discrete wavelengths emitted or absorbed by a specific atom

Step 2: Describe the relationship between energy and wavelength

• Photon energy is related to frequency and wavelength

• Therefore, photons with discrete wavelengths have discrete energies equal to the difference between two energy levels

Step 3: Explain how atomic spectra give evidence for the quantisation of energy 

• Photons arise from electron transitions between energy levels

• This happens when an electron is excited, or de-excited, from one energy level to another, by either emitting or absorbing light of a specific wavelength

• Since atomic spectra are made up of discrete wavelengths, this shows that atoms must contain discrete, or quantised, energy levels 

The diagram shows the electron energy levels in an atom of hydrogen.

Determine the number of possible wavelengths that can be produced from transitions between the n = 4 excited state and the n = 1 ground state.

Answer:

• There are six possible wavelengths that could be produced from the different energy level transitions

• The possible transitions are:

  • n = 4 to n = 3

  • n = 4 to n = 2

  • n = 4 to n = 1

  • n = 3 to n = 2

  • n = 3 to n = 1

  • n = 2 to n = 1

Make sure you learn the definition for a photon: discrete quantity / packet / quantum of electromagnetic energy are all acceptable definitions

Some electron energy levels in atomic hydrogen are shown below.

The longest wavelength produced as a result of electron transitions between two of the energy levels is 4.0 × 10^–6 m.

(a) Draw an arrow to show the transition that would produce:

• A photon of wavelength 4.0 × 10^–6 m. Mark with the letter L.

• The photon with the shortest wavelength. Mark with the letter S.

(b) Calculate the wavelength for the transition giving rise to the shortest wavelength.

Answer:

(a)

• Photon energy and wavelength are inversely proportional

• Therefore, the largest energy change corresponds to the shortest wavelength (line S)

• The smallest energy change corresponds to the longest wavelength (line L)

(b)

Step 1: Write down the equation linking wavelength and energy

$\lambda =\frac{hc}{∆E}=\frac{hc}{E_2-E_1}$

Step 2: Identify the energy levels that give rise to the shortest wavelength

• The shortest wavelength photon will come from a transition between the energy levels that have the largest difference:

  • $E_2=-0.54\mathrm{eV}$

  • $E_1=-3.4\mathrm{eV}$

• Therefore, the greatest possible difference in energy is 

$∆E=E_2-E_1=-0.54-\left(-3.4\right)=2.86\mathrm{eV}$

Step 3: Calculate the wavelength

• To convert from eV to J: multiply by 1.6 × 10⁻¹⁹ J

$\lambda =\frac{\left(6.63\times {10}^{-34}\right)\left(3.0\times {10}^8\right)}{2.86\times \left(1.6\times {10}^{-19}\right)}$

$\lambda$ = 4.347 × 10⁻⁷ m = 435 nm

Light of wavelength 490 nm is incident normally on a surface, as shown in the diagram.

The power of the light is 3.6 mW. The light is completely absorbed by the surface.

Calculate the number of photons incident on the surface in 30 s.

Answer:

Step 1: Write down the known quantities

• Wavelength, λ = 490 nm = 490 × 10⁻⁹ m

• Power, P = 3.6 mW = 3.6 × 10⁻³ W

• Time, t = 30 s

Step 2: Write the equation for photon energy and write in terms of wavelength

 $E=hf\Rightarrow E=\frac{hc}{\lambda }$

Step 3: Calculate the energy of one photon

$E=\frac{hc}{\lambda }=\frac{\left(6.63\times {10}^{-34}\right)\left(3.0\times {10}^8\right)}{490\times {10}^{-9}}=4.06\times {10}^{-19}\mathrm{J}$

Step 4: Calculate the number of photons hitting the surface every second

$\frac{\mathrm{power}\mathrm{of}\mathrm{light}\mathrm{source}}{\mathrm{energy}\mathrm{of}\mathrm{one}\mathrm{photon}}=\frac{3.6\times {10}^{-3}}{4.06\times {10}^{-19}}=8.87\times {10}^{15}{\mathrm{s}}^{-1}$

Step 5: Calculate the number of photons that hit the surface in 30 s

number of photons in 30 s = (8.87 × 10¹⁵) × 30 = 2.7 × 10¹⁷

The values of Planck’s constant and the speed of light will be included in the data booklet; however, it helps to memorise them to speed up calculations!