SATHEE: Modern Physics 1 Question 53

Modern Physics 1 Question 53

49. A hydrogen like atom (atomic number $Z$ ) is in a higher excited state of quantum number $n$ . The excited atom can make a transition to the first excited state by successively emitting two photons of energy $10.2 e V$ and $17.0 e V$ respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $4.25 e V$ and $5.95 e V$ respectively.

$(1994, 6 M)$

Determine the values of $n$ and $Z$ .

(Ionization energy of $H$ -atom $= 13.6 e V$ )

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Answer:

Correct Answer: 49. (a) $r_{n} = \frac{n^{2} h^{2} ε_{0}}{624 π m_{e} e^{2}}$ (b) $n \approx 25$ (c) $0.546 \AA$

Solution:

From the given conditions

$\begin{aligned} E_{n} - E_{2} = (10.2 + 17) e V = 27.2 e V \\ and E_{n} - E_{3} = (4.25 + 5.95) e V = 10.2 e V \end{aligned}$

Eq. (i) - Eq. (ii) gives

$\begin{aligned} E_{3} - E_{2} & = 17.0 e V \\ or Z^{2} (13.6) \frac{1}{4} - \frac{1}{9} & = 17.0 \end{aligned}$

$\begin{array}{ll} \Rightarrow & Z^{2} (13.6) (5 / 36) = 17.0 \\ \Rightarrow & Z^{2} = 9 or Z = 3 \end{array}$

From Eq. (i) $Z^{2}$ (13.6) $\frac{1}{4} - \frac{1}{n^{2}} = 27.2$

$\begin{aligned} or (3)^{2} (13.6) \frac{1}{4} - \frac{1}{n^{2}} = 27.2 \\ or \frac{1}{4} - \frac{1}{n^{2}} = 0.222 \\ or 1 / n^{2} = 0.0278 or n^{2} = 36 \\ ∴ n = 6 \end{aligned}$

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Modern Physics 1 Question 53

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Modern Physics 1 Question 53

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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