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原子物理 ~ 原子・原子核・素粒子

原子・原子核・素粒子

水素原子の線スペクトル

\begin{eqnarray*}
\lambda &=& 3.646 \times 10^{-7} \ \mbox{m} \times \frac{n^2}{n^2 – 2^2} \ (n=3,4,5,6) \\
\\
\frac{1}{\lambda} &=& R \left( \frac{1}{n’^2} – \frac{1}{n^2} \right)
\end{eqnarray*}
\lambda &=& 3.646 \times 10^{-7} \ \mbox{m} \times \frac{n^2}{n^2 – 2^2} \ (n=3,4,5,6)

\frac{1}{\lambda} &=& R \left( \frac{1}{n’^2} – \frac{1}{n^2} \right)

量子条件 (quantum condition)

$$
mvr = n \frac{h}{2\pi}
$$
mvr = n \frac{h}{2\pi}

振動数条件

$$
h \nu = E – E’
$$
h \nu = E – E’

定常状態の電子の円軌道の半径

\begin{eqnarray*}
m \frac{v_n^2}{r_n} &=& k_0 \frac{e^2}{r_n^2} \ (n=1,2,3 \cdots ) \\
\\
r_n &=& \left( \frac{h}{2 \pi} \right) ^2 \frac{n^2}{k_0 me^2} \ (n=1,2,3 \cdots )
\end{eqnarray*}

定常状態の電子のエネルギー

\begin{eqnarray*}
U &=& – k_0 \frac{e^2}{r_n} \\
\\
E_n &=& K + U = \frac{1}{2} m v_n^2 – k_0 \frac{e^2}{r_n} = – k_0 \frac{e^2}{2r_n} \\
\\
E_n &=& – \frac{2 \pi ^2 k_0^2 me^4}{h^2} \cdot \frac{1}{n^2} \ (n=1,2,3 \cdots ) \\
\\
E_n &=& – \frac{2.18 \times 10^{-18}}{n^2} \ \mbox{J} = – \frac{13.6}{n^2} \ \mbox{eV} \ (n=1,2,3 \cdots )
\end{eqnarray*}
U &=& – k_0 \frac{e^2}{r_n}

E_n &=& K + U = \frac{1}{2} m v_n^2 – k_0 \frac{e^2}{r_n} = – k_0 \frac{e^2}{2r_n}

E_n &=& – \frac{2 \pi ^2 k_0^2 me^4}{h^2} \cdot \frac{1}{n^2} \ (n=1,2,3 \cdots )

E_n &=& – \frac{2.18 \times 10^{-18}}{n^2} \ \mbox{J} = – \frac{13.6}{n^2} \ \mbox{eV} \ (n=1,2,3 \cdots )

水素の輝線の波長

\begin{eqnarray*}
h \nu &=& \frac{hc}{\lambda} E_n – E_{n’} \\
\\
\frac{1}{\lambda} &=& \frac{E_n – E_{n’}}{hc} = \frac{2 \pi ^2 k_0^2 me^4}{ch^3} \left( \frac{1}{n’^2} – \frac{1}{n^2} \right) \\
\\
\\
R &=& \frac{2 \pi ^2 k_0^2 me^4}{ch^3}
\end{eqnarray*}
h \nu &=& \frac{hc}{\lambda} E_n – E_{n’}

\frac{1}{\lambda} &=& \frac{E_n – E_{n’}}{hc} = \frac{2 \pi ^2 k_0^2 me^4}{ch^3} \left( \frac{1}{n’^2} – \frac{1}{n^2} \right)

R &=& \frac{2 \pi ^2 k_0^2 me^4}{ch^3}

放射性崩壊 (radioactive decay)

\begin{eqnarray*}
_{88}^{226} \text{Ra} & \to & _{86}^{222} \text{Rn} + _{2}^{4} \text{He} \\
\\
_{81}^{206} \text{Ti} & \to & _{82}^{206} \text{Pb} + \text{e}^- \\
\end{eqnarray*}
_{88}^{226} \text{Ra} & \to & _{86}^{222} \text{Rn} + _{2}^{4} \text{He}

_{81}^{206} \text{Ti} & \to & _{82}^{206} \text{Pb} + \text{e}^-

半減期 (half-life)

$$
N = N_0 \left( \frac{1}{2} \right) ^{\frac{t}{T}}
$$
N = N_0 \left( \frac{1}{2} \right) ^{\frac{t}{T}}

炭素年代測定法

$$
\text{n} + ^{14}\text{N} \to \text{p} + ^{14}\text{C}
$$
\text{n} + ^{14}\text{N} \to \text{p} + ^{14}\text{C}

質量とエネルギーの等価性

\begin{eqnarray*}
E &=& mc^2 \\
\\
m’ &=& \frac{m}{\sqrt{1- \frac{v^2}{c^2}}} \\
\\
E &=& m’ c^2 = \frac{mc^2}{\sqrt{1- \frac{v^2}{c^2}}} \\
\\
p &=& m’ v = \frac{mv}{\sqrt{1 – \frac{v^2}{c^2}}} \\
\\
E &=& \frac{mc^2}{\sqrt{1 – \frac{v^2}{c^2}}} = mc^2 \left( 1 – \frac{v^2}{c^2} \right) ^{-\frac{1}{2}} \simeq mc^2 + \frac{1}{2} mv^2
\end{eqnarray*}
E &=& mc^2

m’ &=& \frac{m}{\sqrt{1- \frac{v^2}{c^2}}}

E &=& m’ c^2 = \frac{mc^2}{\sqrt{1- \frac{v^2}{c^2}}}

p &=& m’ v = \frac{mv}{\sqrt{1 – \frac{v^2}{c^2}}}

E &=& \frac{mc^2}{\sqrt{1 – \frac{v^2}{c^2}}} = mc^2 \left( 1 – \frac{v^2}{c^2} \right) ^{-\frac{1}{2}} \simeq mc^2 + \frac{1}{2} mv^2

原子核の結合エネルギー

\begin{eqnarray*}
\Delta m &=& Zm_\text{p} + Nm_\text{n} – M \\
\\
\Delta m \cdot c^2 &=& Zm_\text{p} c^2 + Nm_\text{n} c^2 – Mc^2
\end{eqnarray*}
\Delta m &=& Zm_\text{p} + Nm_\text{n} – M

\Delta m \cdot c^2 &=& Zm_\text{p} c^2 + Nm_\text{n} c^2 – Mc^2

核融合 (nuclear fusion)

$$
_{1}^{2} \text{H} + _{1}^{3} \text{H} \to _{0}^{1} \text{n} + _{2}^{4} \text{He}
$$
_{1}^{2} \text{H} + _{1}^{3} \text{H} \to _{0}^{1} \text{n} + _{2}^{4} \text{He}

核分裂 (nuclear fission)

$$
_{92}^{235} \text{U} + _{0}^{1} \text{n} \to _{56}^{144} \text{Ba} + _{36}^{89} \text{Kr} + 3 _{0}^{1} \text{n}
$$
_{92}^{235} \text{U} + _{0}^{1} \text{n} \to _{56}^{144} \text{Ba} + _{36}^{89} \text{Kr} + 3 _{0}^{1} \text{n}

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