SATHEE: Modern Physics 1 Question 54

Modern Physics 1 Question 54

50. A particle of charge equal to that of an electron $- e$ , and mass 208 times of the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge $+ 3 e$ . (Take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system, $(1988, 6 M)$

(a) derive an expression for the radius of the $n^{th}$ Bohr orbit.

(b) find the value of $n$ for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.

(Rydberg’s constant $= 1.097 \times 10^{7} m^{- 1}$ )

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Answer:

Correct Answer: 50. (a) $113.74 \AA$ (b) 3

Solution:

If we assume that mass of nucleus $»$ mass of mu-meson, then nucleus will be assumed to be at rest, only mu-meson is revolving round it.

(a) In $n$ th orbit the necessary centripetal force to the mu-meson will be provided by the electrostatic force between the nucleus and the mu-meson.

Hence,

$\frac{m v^{2}}{r} = \frac{1}{4 π ε_{0}} \frac{(Z e) (e)}{r^{2}}$

Further, it is given that Bohr model is applicable to this system also. Hence

Angular momentum in $n^{th}$ orbit $= \frac{n h}{2 π}$

or $m v r = n \frac{h}{2 π}$

We have two unknowns $v$ and $r$ (in $n^{th}$ orbit). After solving these two equations, we get

$r = \frac{n^{2} h^{2} ε_{0}}{Z π m e^{2}}$

Substituting $Z = 3$ and $m = 208 m_{e}$ , we get

$r_{n} = \frac{n^{2} h^{2} ε_{0}}{624 π m_{e} e^{2}}$

(b) The radius of the first Bohrs orbit for the hydrogen atom is $\frac{h^{2} ε_{0}}{π m_{e} e^{2}}$ .

Equating this with the radius calculated in part (a), we get

$n^{2} \approx 624 or n \approx 25$

$\begin{array}{ll} and & potential energy = - \frac{Z e^{2}}{4 π ε_{0} r} \\ ∴ & Total energy E_{n} = \frac{- Z e^{2}}{8 π ε_{0} r} \end{array}$

Substituting value of $r$ , calculated in part (a),

$E_{n} = \frac{1872}{n^{2}} - \frac{m_{e} e^{4}}{8 ε_{0}^{2} h^{2}}$

But $- \frac{m_{e} e^{4}}{8 ε_{0}^{2} h^{2}}$ is the ground state energy of hydrogen atom and hence is equal to $- 13.6 e V$ .

$\begin{array}{ll} ∴ & E_{n} = \frac{- 1872}{n^{2}} (13.6) e V = - \frac{25459.2}{n^{2}} e V \\ ∴ E_{3} - E_{1} = - 25459.2 \frac{1}{9} - \frac{1}{1} = 22630.4 e V \end{array}$

$∴$ The corresponding wavelength,

$λ (in \AA) = \frac{12375}{22630.4} = 0.546 \AA$

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Modern Physics 1 Question 54

Answer:

इंजीनियरिंग की तैयारी

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 54

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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