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:

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?

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

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

5. Prove that the wavelength of light emitted from the transition n=4 to n=2 is 486.1 nm (HINT: $$ e = hf $$ and $$ c = f \lambda $$)

$$ 13.6(\frac{1}{2^2}- \frac{1}{4^2}) = 2.55eV $$

$$ 2.55eV = \frac{hc}{\lambda} $$ therefore $$ \lambda = \frac{hc}{2.55 \times 1.6 \times 10^-19} = 4.875 \times 10^-7 = 488 nm $$