physics-class-12-unit-12-chapter-07-problem-solving-in-quantum-physics-of-atoms-part-2-l-7-9-ng7zlsaxkgk

### <Success style="font-size:25px"> Problem Solving In Quantum Physics Of Atoms Part 2 L-7 </Success>
### <primary style="font-size:24px"> Problem 5 </primary>
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- Scientists observing the absorption spectra from a distant galaxy hypothesized a hydrogen-like atom with the spectral lines matching those of terrestrial hydrogen. In this unique atom, the electrostatic interaction was replaced by gravitational interaction. The 2 electrically neutral interacting species, called ' $e$ ' and ' $p$ ', had masses $F$ times the masses of electron and proton respectively. Find $\mathrm{F}$.

- $ m_e=9.1 \times 10^{-31} \{kg}$

- $ m_p=1.67 \times 10^{-27} \{kg}$

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<footer style = "font-size: 18px"> $\rightarrow$ Solving Problems in 'Quantum Physics Of Atoms' $\rightarrow$ <span style = "color:blue"> Problem 5 </span> $\rightarrow$ Solution 5 $\rightarrow$ Solution 5 </footer>

### <Success style="font-size:25px"> Problem Solving In Quantum Physics Of Atoms Part 2 L-7 </Success>
### <primary style="font-size:24px"> Problem 6 </primary>
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- What is the percentage error in the estimate of the frequency of the $1^{st}$ Balmer line, if the proton is taken to be fixed instead of having a motion about the electron-proton centre of mass?

- **2-Body System**

- $m_p>m_e$

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

- $R = \frac{m_e e^4}{8 \epsilon ^2 ch^3}$

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<footer style = "font-size: 18px">Solution 5 $\rightarrow$ Solution 5 $\rightarrow$ <span style = "color:blue"> Problem 6 </span> $\rightarrow$ Solution 6 $\rightarrow$ Solution 6 </footer>

### <Success style="font-size:25px"> Problem Solving In Quantum Physics Of Atoms Part 2 L-7 </Success>
### <primary style="font-size:24px"> Problem 7 </primary>
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- The emission spectrum associated with de-excitation of hydrogen atoms is recorded using a grating. For transitions associated with the first line of the Balmer series, a first-order maxima is observed at an angle 20 degrees What is the spacing between the slits of the grating?

- $n_f=2 $

- $ \frac{1}{\lambda_{2,3}}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5 R}{36}$

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<footer style = "font-size: 18px">Solution 6 $\rightarrow$ Solution 6 $\rightarrow$ <span style = "color:blue"> Problem 7 </span> $\rightarrow$ Solution 7 $\rightarrow$ Solution 7 </footer>

### <Success style="font-size:25px"> Problem Solving In Quantum Physics Of Atoms Part 2 L-7 </Success>
### <primary style="font-size:24px"> Solution 7 </primary>
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- $d \sin \theta = m \lambda $

- mE II

- m = 1

- $d = \frac{6570 \times 1}{0.34}$

- $\simeq 19320 A$

- $d = 1.93 \mu m$

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<footer style = "font-size: 18px">Problem 7 $\rightarrow$ Solution 7 $\rightarrow$ <span style = "color:blue"> Solution 7 </span> $\rightarrow$ Problem 8 $\rightarrow$ Solution 8 </footer>

### <Success style="font-size:25px"> Problem Solving In Quantum Physics Of Atoms Part 2 L-7 </Success>
### <primary style="font-size:24px"> Problem 8 </primary>
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- Assume a classical collision between an alpha particle and the Au nucleus. Alphaparticles emitted from radium with average velocity of $v=2 \times 10^7 \mathrm{~m} / \mathrm{s}$ are brought to a momentary halt at a distance $d$ from the $\mathrm{Au}$ nucleus. If $Z=79$ for $A u$, find the value of $d$.

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

- $\epsilon_0=8.85 \times 10^{-12} \{F} / \{m}$

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<footer style = "font-size: 18px">Solution 7 $\rightarrow$ Solution 7 $\rightarrow$ <span style = "color:blue"> Problem 8 </span> $\rightarrow$ Solution 8 $\rightarrow$ Solution 8 </footer>