3.5.6.1 So­lu­tion pipee-a

Ques­tion:

Plug the mass of an elec­tron, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9.109 38 10$\POW9,{-31}$ kg, and the rough size of an hy­dro­gen atom, call it $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 10$\POW9,{-10}$ m, into the ex­pres­sion for the ground state ki­netic en­ergy and see how big it is. Note that $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054 57 10$\POW9,{-34}$ J s. Ex­press in units of eV, where one eV equals 1.602 18 10$\POW9,{-19}$ J.

An­swer:

$$E_1 = \frac{\hbar^2\pi^2}{2m\ell_x^2} = \frac{(\mbox{1.054 ... ...mbox{10$\POW9,{-10}$ m})^2} = \mbox{1.506 10$\POW9,{-18}$ J}$$

or 9.4 eV. The true value is about 4.5 eV. That is in the ball park.