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$$c=\lambda f$

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

$f=8.23\times 10^{14}{\mbox{ Hz}}$$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}}$$\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}}$$f={\frac {\Delta E}{h}}$

$c={\frac {\lambda \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}}$$\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}}$$-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}}$$\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}}$$\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 486.1nm (HINT: $e=hf$$e=hf$ and $c=f\lambda$$c=f\lambda$)

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

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