SATHEE: Modern Physics 1 Question 56

Modern Physics 1 Question 56

52. The ionization energy of a hydrogen like Bohr atom is 4 rydberg.

(1984, 4M)

(a) What is the wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state?

(b) What is the radius of the first orbit for this atom?

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Answer:

Correct Answer: 52. $6.6 \times 10^{- 34} J$ -s

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Solution:

(a) 1 rydberg $= 2.2 \times 10^{- 18} J = R h c$

Ionisation energy is given as 4 rydberg

$= 8.8 \times 10^{- 18} J = \frac{8.8 \times 10^{- 18}}{1.6 \times 10^{- 19}} = 55 e V$

$∴$ Energy in first orbit $E_{1} = - 55 e V$

Energy of radiation emitted when electron jumps from first excited state $(n = 2)$ to ground state $(n = 1)$ :

$E_{21} = \frac{E_{1}}{(2)^{2}} - E_{1} = - \frac{3}{4} E_{1} = 41.25 e V$

$∴$ Wavelength of photon emitted in this transition would be

$λ = \frac{12375}{41.25} = 300 \AA$

(b) Let $Z$ be the atomic number of given element. Then

$\begin{aligned} E_{1} & = (- 13.6) (Z^{2}) \\ or & - 55 & = (- 13.6) (Z^{2}) \\ or & Z & \approx 2 \\ Now, as & r & \propto \frac{1}{Z} \end{aligned}$

Radius of first orbit of this atom,

$r_{1} = \frac{r_{H_{1}}}{Z} = \frac{0.529}{2} = 0.2645 \AA$

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Modern Physics 1 Question 56

52. The ionization energy of a hydrogen like Bohr atom is 4 rydberg.

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 56

52. The ionization energy of a hydrogen like Bohr atom is 4 rydberg.

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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