这是一份leeds利兹大学PHYS201501作业代写的成功案例

$$
H=\frac{p^{2}}{2 m_{e}}-V(r)
$$
Making the usual substitution: $\boldsymbol{p}=-i \hbar \nabla$ we get the relevant form of (114):
$$
\left(-\frac{\hbar^{2}}{2 m_{e}} \nabla^{2}-V(r)\right) \psi(\boldsymbol{r})=E \psi(\boldsymbol{r})
$$
It is convenient to use atomic units where the natural unit of length is the Bohr radius: $a_{0} \equiv \hbar^{2} / m e^{2}=0.52910^{-8} \mathrm{~cm}$, and the natural unit of energy is twice the Rydberg constant: $e^{2} / a_{0} \equiv 2 R y=27.2 \mathrm{eV}=4.3610^{-11}$ erg. In these units, $e=\hbar=m=1$.
Equation (119) then takes the form:
$$
\left(\frac{1}{2} \nabla^{2}+E+V(r)\right) \psi(\boldsymbol{r})=0
$$

PHYS201501 COURSE NOTES ：

$$
\frac{1}{2} m v^{2}=\frac{Z e^{2}}{2 r}
$$
For the ground-state:
$$
r \simeq \frac{a_{0}}{Z},
$$
and thus:
$$
v \simeq\left(\frac{Z^{2} e^{2}}{m a_{0}}\right)^{1 / 2}=(Z \alpha) c
$$