SATHEE: Atoms And Nuclei Question 325

Atoms And Nuclei Question 325

Question: If the series limit wavelength of Lyman series for the hydrogen atom is $912 \overset{o}{A},$ , then the series limit wavelength for Balmer series of hydrogen atoms is

Options:

A) $912 \overset{o}{A},$

B) $912 \times 2 \overset{o}{A},$

C) $912 \times 4 \overset{o}{A},$

D) $\frac{912}{2} \overset{o}{A},$

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Answer:

Correct Answer: C

Solution:

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For limiting wavelength of Lyman series $n_{1} = 1,, n_{2} = \infty \frac{1}{λ_{L} = R}$

For limiting wavelength of Balmer series $n_{1} = 1, n_{2} = \infty$

$\frac{1}{λ_{B}} = R (\frac{1}{4}) \Rightarrow λ_{B} = \frac{4}{R}$
$∴ λ_{B} = 4 λ_{1} = 4 \times 912 \overset{o}{A}, .$

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Atoms And Nuclei Question 325

Question: If the series limit wavelength of Lyman series for the hydrogen atom is $912 \overset{o}{A},$ , then the series limit wavelength for Balmer series of hydrogen atoms is

Options:

Answer:

Solution:

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Atoms And Nuclei Question 325

Question: If the series limit wavelength of Lyman series for the hydrogen atom is 912Ao, , then the series limit wavelength for Balmer series of hydrogen atoms is

Options:

Answer:

Solution:

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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Question: If the series limit wavelength of Lyman series for the hydrogen atom is $912 \overset{o}{A},$ , then the series limit wavelength for Balmer series of hydrogen atoms is