Question 39.44: A hydrogen atom in a state having a binding energy (the ener......
A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of 0.85 eV makes a transition to a state with an excitation energy (the difference between the energy of the state and that of the ground state) of 10.2 eV. (a) What is the energy of the photon emitted as a result of the transition? What are the (b) higher quantum number and (c) lower quantum number of the transition producing this emission?
(a) Since E_2 = – 0.85 eV and E_1 = – 13.6 eV + 10.2 eV = – 3.4 eV, the photon energy is
E_{\text {photon }}=E_2-E_1=-0.85 \,eV -(-3.4 \, eV )=2.6 \,eV .
(b) From
E_2-E_1=(-13.6 \,eV )\left(\frac{1}{n_2^2}-\frac{1}{n_1^2} \right)\,=2.6 \,eV
we obtain
\frac{1}{n_2^2}-\frac{1}{n_1^2}=\frac{2.6 eV }{13.6 eV } \approx-\frac{3}{16}=\frac{1}{4^2}-\frac{1}{2^2} .
Thus, n_2 = 4 and n_1 = 2. So the transition is from the n = 4 state to the n = 2 state. One can easily verify this by inspecting the energy level diagram of Fig. 39-18. Thus, the higher quantum number is n_2 = 4.
(c) The lower quantum number is n_1 = 2.