$n=3$

$n=2$

$n=1$

$\begin{gathered}E_n \sim-\frac{W}{n^2} \ n \in I \ n_f=2\end{gathered}$

$n_f=2$

$n_i=3,4,5$

$E_n \sim \frac{1}{n^2} E_n \sim h \nu \sim \frac{h c}{\lambda}$

$\frac{1}{\lambda_{n_f, n_i}}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

$R$ = $Rydberg$ $constant$

$\begin{aligned} \frac{1}{\lambda_{2,3}} & =R\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \ & =5 R / 36\end{aligned}$

$\frac{1}{\lambda_{2,4}}=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3 R}{16}$

$\begin{aligned} \frac{\lambda_{2,4}}{\lambda_{2,3}} & =\frac{5 R}{36} \times \frac{16}{3 R}=\frac{20}{27} \ & =20 / 27\end{aligned}$

$\lambda_{2,4}={6550} \times \frac{20}{27}= 4850 \mathring{A}$

$\quad \quad \quad \quad \downarrow$ $\quad \quad \quad \quad \downarrow$

$\quad \quad \quad \quad H_\alpha$ $\quad \quad \quad H_\beta$

$\frac{1}{\lambda_{n_f, n_i}}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Balmer: $n_f=2, \quad n_i=3 \rightarrow$ Straight line

Lyman: $\quad n_f=1, \quad n_i=2$

$\frac{1}{\lambda_{2,3}}=\frac{5 R}{36}$

$\frac{1}{\lambda_{1,2}}=\frac{3 R}{4}$

$\begin{aligned} & \lambda_{1,2}=\frac{5}{27} \times \lambda_{2,3} \ & \approx 1210 \mathring{A}\end{aligned}$

Assume that the velocity of an electron in the first Bohr orbit of hydrogen can be defined by Bohr's angular momentum postulate.

Find its velocity and comment if the electron is relativistic.

$e=-1.6 \times 10^{-19} \mathrm{C} .$

$\varepsilon_0=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$

$h$ = $6.626 \times 10^{-34}$

${m^2}{kg}/{s}$.or J-s

$<\psi_1 |\hat{\frac{p}{m}}| \psi_1>$

$Later$

$L=m vr=n \hbar = \frac{nh}{2\pi}$

$n=1$ , $r=?$

$Centripetal$ $force$ = $Columb$ $force$

$\frac{m v^2}{r}=\frac{e^2}{4 \pi \varepsilon_0 \gamma^2}$

$m v^2 r=\frac{e^2}{4 \pi \varepsilon_0}$

$r=\frac{e^2}{4 \pi \varepsilon_0 m v^2}$

$m v r=\frac{e^2}{4 \pi \varepsilon_0 v}=\frac{n h}{2 \pi}$

$v=\frac{e^2}{\varepsilon_0 h} \rightarrow$ $0.219 \times 10^{7} m/sec$

$\approx \frac{3 c}{137} \leqslant c$

Non-relavistic, $v \ll C$

Bohr radius

$v=\frac{n h}{2 \pi mr}$

Collisions knock off the lowest energy electron in a helium atom, and raise the remaining electron to various excited states.

If the resultant $\mathrm{He}^{+}$ion is irradiated with a broadband wavelength source, then find the largest wavelength that will be absorbed.

$h$ = $6.626 \times 10^{-34}$

${m^2}{kg}/{s}$ or $Js=4.14 \times 10^{-15} eV-s$

$c=3 \times 10^8 m / s$

$\mathrm{He}: \mathrm{1s}^2$

$\mathrm{He}^{+} \rightarrow \text { Isoelectronic }$

${H} {-atom }$

$E_n=Ƶ^{2} R\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]$

$\quad \quad \quad \uparrow$

$\text { Ƶ = of protons }$

$\frac{1}{\lambda_{n_f, n_i}}=\frac{2^2 \cdot(13 \cdot 6)_e V}{h c} x$

$\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$

$\frac{1}{\lambda_{n_{f}, n_i}}=\frac{4.0 \times 13.6 \mathrm{eV}}{4.14 \times 10^{-15} \times 3 \times 10^8} \times \frac{3}{4} \mathrm{~m}^{-1}$

$\lambda_{\max } \approx 300 \mathring{A}$

Second

$R =109678 {cm}^{-1}$

$\cong 10967800 {m}^{-1}$

$\frac{1}{\lambda_{n_f, n_i}}=4 R \cdot \frac{3}{4}=3 R$

$\lambda_{1,2}=\frac{1}{3 R} \cong 300 \mathring{A}$