SATHEE: Modern Physics 1 Question 30

Modern Physics 1 Question 30

27. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{π} \times 10^{11} H z$ . Then the moment of inertia of $C O$ molecule about its centre of mass is close to (Take $h = 2 π \times 10^{- 34} J - s)$

(a) $2.76 \times 10^{- 46} k g - m^{2}$

(b) $1.87 \times 10^{- 46} k g - m^{2}$

(d) $1.17 \times 10^{- 47} k g - m^{2}$

(2010)

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Solution:

$h v = K_{2} - K_{1} = \frac{3 h^{2}}{8 π^{2} I}$

$\begin{aligned} ∴ I = \frac{3 h}{8 π^{2} f} & = \frac{3 \times 2 π \times 10^{- 34} \times π}{8 \times π^{2} \times 4 \times 10^{11}} \\ = 1.87 \times 10^{- 46} k g m^{2} \end{aligned}$

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Modern Physics 1 Question 30

27. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{π} \times 10^{11} H z$ . Then the moment of inertia of $C O$ molecule about its centre of mass is close to (Take $h = 2 π \times 10^{- 34} J - s)$

Solution:

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चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Modern Physics 1 Question 30

27. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to 4π×1011Hz. Then the moment of inertia of CO molecule about its centre of mass is close to (Take h=2π×10−34J−s)

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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27. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{π} \times 10^{11} H z$ . Then the moment of inertia of $C O$ molecule about its centre of mass is close to (Take $h = 2 π \times 10^{- 34} J - s)$