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

Bluegreen

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?
$\Delta E={\frac {ch}{\lambda _{5,2}}}{\frac {ch}{\lambda _{7,2}}}=ch\left({\frac {1}{\lambda _{5,2}}}{\frac {1}{\lambda _{7,2}}}\right)=3\times 10^{8}\times 6.63\times 10^{34}\left({\frac {1}{434.1\times 10^{9}}}{\frac {1}{397\times 10^{9}}}\right)=4.28\times 10^{20}{\mbox{ J}}$
$\lambda ={\frac {ch}{\Delta E}}={\frac {3\times 10^{8}\times 6.63\times 10^{34}}{4.28\times 10^{20}}}=4.65{\mbox{ }}\mu {\mbox{m}}$
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 $\lambda ={\frac {hc}{2.55\times 1.6\times 10^{}19}}=4.875\times 10^{}7=488nm$