原子物理 ~ 電子と光

$$\begin{eqnarray*} \frac{e}{m} &=& \frac{v}{Br} \\ \\ &=& \frac{2V}{B^2 r^2} \\ \\ \\ \frac{e}{m} &=& 1.75881962 \times 10^{11} \ \mbox{C/kg} \end{eqnarray*}$$

\frac{e}{m} &=& \frac{v}{Br}

&=& \frac{2V}{B^2 r^2}

\frac{e}{m} &=& 1.75881962 \times 10^{11} \ \mbox{C/kg}

$$\begin{eqnarray*} E &=& h\nu \\ \\ h &=& 6.62607004 \times 10^{-34} \ \mbox{J}\cdot \mbox{s} \ \text{(プランク定数)} \end{eqnarray*}$$

E &=& h\nu

h &=& 6.62607004 \times 10^{-34} \ \mbox{J}\cdot \mbox{s} \ \text{(プランク定数)}

$$\begin{eqnarray*} K_0 = eV_0 = h \nu – W \end{eqnarray*}$$

K_0 = eV_0 = h \nu – W

$$\begin{eqnarray*} eV &=& h \nu _0 = \frac{hc}{\lambda _0} \\ \\ \lambda _0 &=& \frac{hc}{eV} \end{eqnarray*}$$

eV &=& h \nu _0 = \frac{hc}{\lambda _0}

\lambda _0 &=& \frac{hc}{eV}

$$2d \sin \theta = n \lambda$$

2d \sin \theta = n \lambda

$$p = \frac{h \nu}{c} = \frac{h}{\lambda}$$

p = \frac{h \nu}{c} = \frac{h}{\lambda}

$$\lambda ' = \lambda + \frac{h}{mc} (1 – \cos \theta)$$

\lambda ' = \lambda + \frac{h}{mc} (1 – \cos \theta)

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

\lambda = \frac{h}{p} = \frac{h}{mv}