The diagram shows an energy-level diagram for a hydrogen atom - AQA - A-Level Physics - Question 13 - 2018 - Paper 1

To determine the number of different wavelengths that can be emitted when the hydrogen atoms de-excite, we first need to calculate the energy of the incident electrons in electronvolts (eV).

The relationship between energy in joules (J) and electronvolts (eV) is given by:

$E( ext{eV}) = \frac{E(J)}{1.6 \times 10^{-19} \text{ J/eV}}$

For the given kinetic energy of electrons:

$E = 2.0 \times 10^{-18} \text{ J}$

Substituting this into the equation gives us:

$E = \frac{2.0 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 12.5 \text{ eV}$

Next, we compare this energy with the energy levels of hydrogen atom:

• Ground state: -13.6 eV
• First excited state: -3.4 eV
• Second excited state: -1.51 eV
• Third excited state: -0.85 eV
• Fourth excited state: -0.54 eV

The incident energy of 12.5 eV can promote the electron from the ground state (-13.6 eV) to the higher states. The energy transitions (and hence the emitted photons) will occur as follows:

1. From -13.6 eV to -3.4 eV
2. From -13.6 eV to -1.51 eV
3. From -13.6 eV to -0.85 eV
4. From -13.6 eV to -0.54 eV

Calculating all possible transitions, we find:

• From -3.4 eV to -1.51 eV
• From -3.4 eV to -0.85 eV
• From -3.4 eV to -0.54 eV
• From -1.51 eV to -0.85 eV
• From -1.51 eV to -0.54 eV
• From -0.85 eV to -0.54 eV

Thus, the total number of different wavelengths (emitted photons) that can be observed is 6. Therefore, the answer is:

C: 6