4.3.2.3 So­lu­tion hydb-c

Ques­tion:

Plug num­bers into the generic ex­pres­sion for the en­ergy eigen­val­ues,

$$E_n = - \frac{\hbar^2}{2m_{\rm e}a_0^2} \frac 1{n^2},$$

where $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $4\pi\epsilon_0\hbar^2$$\raisebox{.5pt}{$/$}$​${m_{\rm e}}e^2$, to find the ground state en­ergy. Ex­press in eV, where 1 eV equals 1.602 2 10$\POW9,{-19}$ J. Val­ues for the phys­i­cal con­stants can be found at the start of this sec­tion and in the no­ta­tions sec­tion.

An­swer:

First ver­ify the Bohr ra­dius

$$a_0 = \frac{4\pi\;\mbox{8.854 10$\POW9,{-12}$ C$\POW9,{2}$/... ...2 10$\POW9,{-19}$ C})^2} = \mbox{0.529 2 10$\POW9,{-10}$ m}$$

Next, tak­ing $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for the ground state,

$$E_1 = - \frac{(\mbox{1.054 6 10$\POW9,{-34}$J s})^2} {2\;\m... ...8}$ J} \frac{\mbox{1 eV}}{\mbox{1.602 2 10$\POW9,{-19}$ J}}$$

which gives $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$13.61 eV