Question 1.10.17: Use the uncertainty principle to estimate: (a) the ground st...

(a) According to the uncertainty principle, the electron’s momentum and the radius of its orbit are related by rp\thicksim \hbar ;  hence  p\thicksim \hbar/r.  To find the ground state radius, we simply need to minimize the electron–proton energy

E(r)=\frac{p^2}{2m_e}-\frac{e^2}{4\pi \varepsilon _0r}=\frac{\hbar ^2}{2m_er^2}-\frac{e^2}{4\pi \varepsilon _0r}              (1.236)

with respect to r:

0=\frac{dE}{dr}=-\frac{\hbar^2}{m_er_0^3}+\frac{e^2}{4\pi \varepsilon _0r_0^2}.               (1.237)

This leads to the Bohr radius

r_0=\frac{4\pi \varepsilon _0\hbar ^2}{m_ee^2}=0.053  nm.                (1.238)

(b) Inserting (1.238) into (1.236), we obtain the Bohr energy:

E(r_0)=\frac{\hbar^2}{2mr_0^2}-\frac{e^2}{4\pi \varepsilon _0r_0}=-\frac{m_e}{2\hbar ^2}\left(\frac{e^2}{4\pi \varepsilon _0} \right)^2=-13.6  eV.              (1.239)

The results obtained for r_0  and E(r_0)  , as shown in (1.238) and (1.239), are indeed impressively accurate given the crudeness of the approximation.