The diagram shows some energy levels of a hydrogen atom.
(a) An electron moves from the $n = 3$ level to the $n = 2$ level. Calculate the wavelength of the photon emitted. [3]
(b) State the number of different photon energies that could be emitted by atoms excited to the $n = 4$ level. [1]
(c) A free electron with kinetic energy 12.5 eV collides with a hydrogen atom in its ground state. State the highest level to which the atom can be excited. [1]
- (a) $\Delta E = 3.40 - 1.51 = 1.89$ eV $= 1.89\times1.60\times10^{-19} = 3.02\times10^{-19}$ J(1)
- (a) $\lambda = \dfrac{hc}{\Delta E} = \dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{3.02\times10^{-19}}$(1)
- (a) $6.6 \times 10^{-7}$ m(1)
- (b) 6(1)
- (c) $n = 3$ (needs 12.09 eV; $n = 4$ would need 12.75 eV)(1)
- Answer: (a) $6.6\times10^{-7}$ m (b) 6 (c) $n = 3$
- Examiner tip: A colliding electron can give part of its energy, but a photon must be absorbed whole — read whether it is an electron or a photon.