# 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.

${\displaystyle c=\lambda f}$

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

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

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

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

This time, let's derive a general formula:

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

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

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

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

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

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

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

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

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

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

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