# A-level Physics (Advancing Physics)/Energy Levels/Worked Solutions

The following table gives the wavelengths of light given off when electrons change between the energy levels in hydrogen as described in the first row:

 Transition of n Wavelength (nm) Colour 3→2 4→2 5→2 6→2 7→2 8→2 9→2 ∞→2 656.3 486.1 434.1 410.2 397.0 388.9 383.5 364.6 Red Blue-green Violet Violet (Ultraviolet) (Ultraviolet) (Ultraviolet) (Ultraviolet)

1. Calculate the potential energy of an electron at level n=2.

$c = \lambda f$

$3 \times 10^8 = 364.6 \times 10^{-9} \times f$

$f = 8.23 \times 10^{14}\mbox{ Hz}$

$\Delta E = -hf = -6.63 \times 10^{-34} \times 8.23 \times 10^{14} = -5.46 \times 10^{-19}\mbox{ J} = -3.41\mbox{ eV}$

2. Calculate the difference in potential energy between levels n=2 and n=3.

This time, let's derive a general formula:

$f = \frac{\Delta E}{h}$

$c = \frac{\lambda \Delta E}{h}$

$\Delta E = \frac{ch}{\lambda} = \frac{3 \times 10^8 \times 6.63 \times 10^{-34}}{656.3 \times 10^{-9}} = 3.03 \times 10^{-19}\mbox{ J} = 1.89\mbox{ eV}$

3. What is the potential energy of an electron at level n=3?

$-3.41 + 1.89 = -1.52\mbox{ eV}$

4. If an electron were to jump from n=7 to n=5, what would the wavelength of the photon given off be?

$\Delta E = \frac{ch}{\lambda_{5,2}} - \frac{ch}{\lambda_{7,2}} = ch\left ( \frac{1}{\lambda_{5,2}} - \frac{1}{\lambda_{7,2}}\right ) = 3 \times 10^8 \times 6.63 \times 10^{-34} \left ( \frac{1}{434.1 \times 10^{-9}} - \frac{1}{397 \times 10^{-9}}\right ) = -4.28 \times 10^{-20}\mbox{ J}$

$\lambda = \frac{-ch}{\Delta E} = \frac{3 \times 10^8 \times 6.63 \times 10^{-34}}{4.28 \times 10^{-20}} = 4.65\mbox{ }\mu\mbox{m}$