Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
Question
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Chapter 41, Problem 10P

(a)

To determine

Prove that the period of electron is T=n3t0 and calculate the numerical value of t0.

(a)

Expert Solution
Check Mark

Answer to Problem 10P

It is proved that T=n3t0 and the numerical value of t0 is 152 as.

Explanation of Solution

Write the equation for the angular momentum of electron.

    L=mevr

Here, L is the angular momentum of electron, me is the mass of electron, v is the linear speed of electron, and r is the radius of orbit.

Write the quantization condition for L.

    L=n

Here, n is the principal quantum number, and is the reduced Plank’s s constant.

Equate the right hand sides of above two equations.

    mevr=n

Rewrite the above expression in terms of v.

    v=nmer

Write the equation for time period of orbital motion.

    T=2πrv

Here, T is the time period of orbital motion.

Rewrite the above expression by substituting 2πrv for T.

    T=2πr(nmer)=2πrmern

Write the equation to find the orbital radius.

    r=n22mekee2

Here, r is the orbital radius, ke is the Coulomb’s constant, and e is the magnitude of charge of electron.

Rewrite the expression for T by substituting the above relation.

    T=2πrme(n22mekee2)n=2π3meke2e4n3

From dimensional analysis, it can be seen that the whole term 2π3meke2e4 has the dimension of time. So it can be treated as t0.

  T=2π3meke2e4n3T=t0n3                                                                                                           (I)

Thus, it is it is proved that T=n3t0.

Conclusion:

Substitute 3.14 for π, 1.054 6×1034 Js for , 9.11×1031kg for me, 8.99×109 Nm2/C2 for ke, 1.602×1019C for e, and 1 for n in equation (I) to find T.

    T=2(3.14)(1.054 6×1034 Js)3(9.11×1031 kg)(8.99×109 Nm2/C2)2(1.602×1019 C)4 (1)3=7.36×101024849.4×1089s=1.52×1016 s(1as1018 s)=152as

Therefore, it is proved that T=n3t0 and the numerical value of t0 is 152 as.

(b)

To determine

Number of revolutions made by electron in the excited state.

(b)

Expert Solution
Check Mark

Answer to Problem 10P

Number of revolutions made by electron in the excited state is 8.23×109.

Explanation of Solution

Write the equation to find the number of revolutions made by electron in the excited state.

    N=tlifeT

Here, N is the number of revolutions made by electron in the excited state and the tlife is the lifetime in excited state.

Conclusion:

Substitute 3.14 for π, 1.054 6×1034 Js for , 9.11×1031kg for me, 8.99×109 Nm2/C2 for ke, 1.602×1019C for e, and 2 for n in equation (I) to find T for excited state.

  T=2(3.14)(1.054 6×1034 Js)3(9.11×1031 kg)(8.99×109 Nm2/C2)2(1.602×1019 C)4 (2)3=7.36×101024849.4×1089(8)s=1.22×1015s

Substitute 10μs for tlife and 1.22×1015s for T in the equation for N.

  N=10μs(106s1μs)1.22×1015s=8.23×109

Therefore, the number of revolutions made by electron in the excited state is 8.23×109.

(c)

To determine

Check whether is it possible to treat the electron in n=2 as “living for a long time” on considering the period of revolution of electron as a year.

(c)

Expert Solution
Check Mark

Answer to Problem 10P

On comparing the period of revolution electron with that of Sun, it is correct that electron in n=2 is “living for a long time”.

Explanation of Solution

It is found that in part (b), T for electron at n=2 is 1.22×1015s. Electron makes 8.23×109 revolutions in 16μs. If it is assumed that the period of electron as one year, electron stays at n=2 for 8.23×109 years. It is a very big value compared to the period of revolution of Sun.

Conclusion:

Therefore, on comparing the period of revolution electron with that of Sun, it is correct that electron in n=2 is “living for a long time”.

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Students have asked these similar questions
An electron is in the nth Bohr orbit of the hydrogen atom. (a) Show that the period of the electron is T = n3t0 and determine the numerical value of t0. (b) On average, an electron remains in the n = 2 orbit for approximately 10 ms before it jumps down to the n = 1 (ground-state) orbit. How many revolutions does the electron make in the excited state? (c) Define the period of one revolution as an electron year, analogous to an Earth year being the period of the Earth’s motion around the Sun. Explain whether we should think of the electron in the n = 2 orbit as “living for a long time.”
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