Дано: Δx Δp_x ≥ ℏ Δx = r n = 1 Z = 1 t = h / (2π) Требуется найти: E_min

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