Answer:
a) 10.2 eV
b) 122 nm
Explanation:
a) First we must obtain the energy for each of the states, which is given by the following formula:
[tex]E_{n}=\frac{-13.6 eV}{n^2}[/tex]
So, we have:
[tex]E_{1}=\frac{-13.6 eV}{1^2}=-13.6 eV\\E_{2}=\frac{-13.6 eV}{2^2}=-3.4 eV[/tex]
Now we find the energy that the electron loses when it falls from state 2 to state 1, this is the energy carried away by the emitted photon.
[tex]E_{2}-E_{1}=-3.4eV-(-13.6eV)=10.2eV[/tex]
b) Using the Planck – Einstein relation, we can calculate the wavelength of the photon:
[tex]E=h\nu[/tex]
Where E is the photon energy, h the Planck constant and [tex]\nu[/tex] the frequency.
Recall that [tex]\nu=\frac{c}{\lambda}[/tex], Rewriting for [tex]\lambda[/tex]:
[tex]E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.13*10^{-15}eV)(3*10^8\frac{m}{s})}{10.2eV}=1.22*10^{-7}m[/tex]
Recall that [tex]1 m=10^9nm[/tex], So:
[tex]1.22*10^{-7}m*\frac{10^9nm}{1m}=122nm[/tex]
