physics-class-12-unit-12-chapter-06-problem-solvingin-quantum-physics-of-atoms-part-1-ztt2akahw2s

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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<footer style = "font-size: 18px"> $\rightarrow$ Problem Solving in Quantum Physics of Atoms $\rightarrow$ <span style = "color:blue"> Public Domain Image </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution-1 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- The first line of the balmer series of the hydrogen atom has wavelength, $\lambda \sim 6550 \mathring{A}$. Find the wavelength of the second line of larger frequency.

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<footer style = "font-size: 18px">Problem Solving in Quantum Physics of Atoms $\rightarrow$ Public Domain Image $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution-1 $\rightarrow$ Solution-1 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- $n=3$

- $n=2$

- $n=1$

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

- $ 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} $

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

- $R$ = $Rydberg$ $constant$

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<footer style = "font-size: 18px">Public Domain Image $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution-1 </span> $\rightarrow$ Solution-1 $\rightarrow$ Solution-1 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- $\begin{aligned} \frac{\lambda_{2,4}}{\lambda_{2,3}} & =\frac{5 R}{36} \times \frac{16}{3 R}=\frac{20}{27} \\ & =20 / 27\end{aligned}$

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

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

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

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<footer style = "font-size: 18px">Solution-1 $\rightarrow$ Solution-1 $\rightarrow$ <span style = "color:blue"> Solution-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution-2 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- $\frac{1}{\lambda_{n_f, n_i}}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

- 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} $

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

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

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<footer style = "font-size: 18px">Solution-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution-2 </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution-3 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- 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

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution-2 $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution-3 $\rightarrow$ Solution-3 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- $<\psi_1 |\hat{\frac{p}{m}}| \psi_1>$

- $ Later $

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

- $ n=1 $ , $ r=?$

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

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<footer style = "font-size: 18px">Solution-2 $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution-3 </span> $\rightarrow$ Solution-3 $\rightarrow$ Solution-3 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- $ \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 $

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution-3 $\rightarrow$ <span style = "color:blue"> Solution-3 </span> $\rightarrow$ Solution-3 $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Problem Solving in Quantum Physics of Atoms Part 1 </Success>
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- 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 $

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<footer style = "font-size: 18px">Solution-3 $\rightarrow$ Solution-3 $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution-4 $\rightarrow$ Solution-4 </footer>