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

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

$c=\lambda f$

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

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

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

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

This time, let's derive a general formula:

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

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

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

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

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

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

$\mathrm{\Delta }E=\frac{ch}{{\lambda }_{5,2}}-\frac{ch}{{\lambda }_{7,2}}=ch(\frac{1}{{\lambda }_{5,2}}-\frac{1}{{\lambda }_{7,2}})=3\times 10^8\times 6.63\times 10^{-34}(\frac{1}{434.1\times 10^{-9}}-\frac{1}{397\times 10^{-9}})=-4.28\times 10^{-20}\text{ J}$

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

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