Условие:
| Дано | Решение |
|---|---|
| $ \begin{array}{l} \Delta x \Delta p{\lambda} \geqslant \hbar | |
| \Delta x=r \end{array} $ | $\Delta x \Delta p{x} \geqslant \hbar, \quad \frac{\Delta p{x}}{p{x}} \approx 1, \quad p{x}=\Delta p{x}=\frac{\hbar}{\Delta x}=\frac{\hbar}{r}$, |
| $ \begin{array}{l} n=1 | |
| Z=1 \end{array} $ | $E=T+\Pi=\frac{p^{2}}{2 m}+\left(-\frac{e^{2}}{4 \pi \varepsilon{0} r}\right), \quad E=\frac{\hbar^{2}}{2 m r^{2}}-\frac{e^{2}}{4 \pi \varepsilon{0} r}$, |
| E{\text {min }-?} | |
| $ r{\min } \frac{\mathrm{d} E}{\mathrm{~d} r}=0, $ | $\frac{\mathrm{d} E}{\mathrm{~d} r}=-\frac{\hbar^{2}}{m r^{3}}+\frac{e^{2}}{4 \pi \varepsilon{0} r^{2}}, \quad \frac{1}{r^{2}}\left(\frac{e^{2}}{4 \pi \varepsilon{0}}-\frac{\hbar^{2}}{m r}\right)=0$, |
| $ r{\min }=\frac{4 \pi \varepsilon{0} \hbar^{2}}{m e^{2}}, $ | $ E{\min }=\frac{\hbar^{2}}{2 m r{\min }^{2}}-\frac{e^{2}}{4 \pi \varepsilon{0} r{\min }}=-\frac{m e^{4}}{2\left(4 \pi \varepsilon{0}\right)^{2} \hbar^{2}}, $ |
| $t=\frac{h}{2 \pi}$, | $ E{\min }=-\frac{m e^{4}}{8 h^{2} \varepsilon{0}^{2}} $ |
| Ombem E{\text {min }} | $\frac{m e^{4}}{8 h^{2} \varepsilon_{0}^{2}}=-13,6$ эВ. |