Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 42, Problem 91CP

(a)

To determine

Show that the frequency of the light emitted when electron moves from n state to n1 state is f=(2π2meke2e4h3)2n1n2(n1)2.

(a)

Expert Solution
Check Mark

Answer to Problem 91CP

It is shown that the frequency of the light emitted when electron moves from n state to n1 state is f=(2π2meke2e4h3)2n1n2(n1)2_.

Explanation of Solution

Write the expression for the allowed energies of an atom.

    En=kee22a0(1n2)                                                                                             (I)

Here, En is the possible energies, ke is the Coulomb’s constant, e is the charge of electron, a0 is the Bohr radius, and n is the principle quantum number corresponding to each energy level.

Write the expression for Bohr radius.

    a0=h24π2mekee2                                                                                               (II)

Here, me is the mass of electron, and h is the Planck’s constant.

Use expression (II) in (I).

    En=kee224π2mekee2h2(1n2)=2π2meke2e4h2(1n2)                                                                                (III)

Here the transition of electron takes place from n to n1 state.

Write the expression for the energy released during a transition from n to n1 state.

    ΔE=hf                                                                                                (IV)

Here, ΔE is the energy released during a transition from n to n1 state, and f is the frequency of the photon.

Use expression (III) to find the energy released during a transition from n to n1 state.

    ΔE=2π2meke2e4h2(1(n1)21n2)                                                                        (V)

Simplify expression (V).

    ΔE=2π2meke2e4h2(1(n1)21n2)=2π2meke2e4h2(n2(n22n+1)n2(n1)2)=2π2meke2e4h2(2n1n2(n1)2)                                                                       (VI)

Equate expression (VI) and (IV) and solve for f.

    hf=2π2meke2e4h2(2n1n2(n1)2)f=(2π2meke2e4h3)(2n1n2(n1)2)

Conclusion:

Therefore, it is shown that the frequency of the light emitted when electron moves from n state to n1 state is f=(2π2meke2e4h3)2n1n2(n1)2_.

(b)

To determine

Show that when n, the expression for frequency varies as 1n3 and also that it reduces to the classical limit.

(b)

Expert Solution
Check Mark

Answer to Problem 91CP

It is shown that when n, the expression for frequency varies as 1n3 and also that it reduces to the classical limit.

Explanation of Solution

Write the expression for the frequency of light emitted during a transition from n to n1 state.

    f=(2π2meke2e4h3)(2n1n2(n1)2)                                                                   (VII)

When n, expression (VII) reduces to simpler form.

    f=(2π2meke2e4h3)(2n3)=4π2meke2e4h3n3                                                                                   (VIII)

Write the expression for the classical frequency of electron.

    f=v2πr                                                                                                          (IX)

Here, v is the speed of electron, and r is the radius of the circular path followed by electron.

Write the expression for velocity of electron.

    v=kee2mer                                                                                                   (X)

Here, me is the mass of electron.

Use expression (X) in (IX).

    f=12πrkee2mer=kee24π2mer3                                                                                                 (XI)

Rewrite expression (XI) in terms of the radius of the nth orbit of the atom.

    f=kee24π2mern3                                                                                                  (XII)

Here, rn is the radius of the nth orbit of the atom.

Write the expression for the radius of the nth orbit of the atom.

    rn=n2h24π2mekee2                                                                                                     (XIII)

Here, n is the principle quantum number corresponding to the nth orbit of the atom, and h is the Planck’s constant.

Use expression (XIII) in (XII) and solve for f.

    f=kee24π2me(4π2mekee2n2h2)3=(4π2)2me2ke4e8n6h6=4π2meke2e4h3n3                                                                           (XIV)

The classical frequency of the electron undergoing a transition from n to n1 state is 4π2meke2e4h3n3.

That is it is proven that as the value of n approaches larger values, the frequency tends to classical limit. Also it is proven that as n, the frequency varies as 1n3.

Conclusion:

Therefore, it is shown that when n, the expression for frequency varies as 1n3 and also that it reduces to the classical limit.

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Chapter 42 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 42 - Prob. 6OQCh. 42 - Prob. 7OQCh. 42 - Prob. 8OQCh. 42 - Prob. 9OQCh. 42 - Prob. 10OQCh. 42 - Prob. 11OQCh. 42 - Prob. 12OQCh. 42 - Prob. 13OQCh. 42 - Prob. 14OQCh. 42 - Prob. 15OQCh. 42 - Prob. 1CQCh. 42 - Prob. 2CQCh. 42 - Prob. 3CQCh. 42 - Prob. 4CQCh. 42 - Prob. 5CQCh. 42 - Prob. 6CQCh. 42 - Prob. 7CQCh. 42 - Prob. 8CQCh. 42 - Prob. 9CQCh. 42 - Prob. 10CQCh. 42 - Prob. 11CQCh. 42 - Prob. 12CQCh. 42 - Prob. 1PCh. 42 - Prob. 2PCh. 42 - Prob. 3PCh. 42 - Prob. 4PCh. 42 - Prob. 5PCh. 42 - Prob. 6PCh. 42 - Prob. 7PCh. 42 - Prob. 8PCh. 42 - Prob. 9PCh. 42 - Prob. 10PCh. 42 - Prob. 11PCh. 42 - Prob. 12PCh. 42 - Prob. 13PCh. 42 - Prob. 14PCh. 42 - Prob. 15PCh. 42 - Prob. 16PCh. 42 - Prob. 17PCh. 42 - Prob. 18PCh. 42 - Prob. 19PCh. 42 - Prob. 20PCh. 42 - Prob. 21PCh. 42 - Prob. 23PCh. 42 - Prob. 24PCh. 42 - Prob. 25PCh. 42 - Prob. 26PCh. 42 - Prob. 27PCh. 42 - Prob. 28PCh. 42 - Prob. 29PCh. 42 - Prob. 30PCh. 42 - Prob. 31PCh. 42 - Prob. 32PCh. 42 - Prob. 33PCh. 42 - Prob. 34PCh. 42 - Prob. 35PCh. 42 - Prob. 36PCh. 42 - Prob. 37PCh. 42 - Prob. 38PCh. 42 - Prob. 39PCh. 42 - Prob. 40PCh. 42 - Prob. 41PCh. 42 - Prob. 43PCh. 42 - Prob. 44PCh. 42 - Prob. 45PCh. 42 - Prob. 46PCh. 42 - Prob. 47PCh. 42 - Prob. 48PCh. 42 - Prob. 49PCh. 42 - Prob. 50PCh. 42 - Prob. 51PCh. 42 - Prob. 52PCh. 42 - Prob. 53PCh. 42 - Prob. 54PCh. 42 - Prob. 55PCh. 42 - Prob. 56PCh. 42 - Prob. 57PCh. 42 - Prob. 58PCh. 42 - Prob. 59PCh. 42 - Prob. 60PCh. 42 - Prob. 61PCh. 42 - Prob. 62PCh. 42 - Prob. 63PCh. 42 - Prob. 64PCh. 42 - Prob. 65APCh. 42 - Prob. 66APCh. 42 - Prob. 67APCh. 42 - Prob. 68APCh. 42 - Prob. 69APCh. 42 - Prob. 70APCh. 42 - Prob. 71APCh. 42 - Prob. 72APCh. 42 - Prob. 73APCh. 42 - Prob. 74APCh. 42 - Prob. 75APCh. 42 - Prob. 76APCh. 42 - Prob. 77APCh. 42 - Prob. 78APCh. 42 - Prob. 79APCh. 42 - Prob. 80APCh. 42 - Prob. 81APCh. 42 - Prob. 82APCh. 42 - Prob. 83APCh. 42 - Prob. 84APCh. 42 - Prob. 85APCh. 42 - Prob. 86APCh. 42 - Prob. 87APCh. 42 - Prob. 88APCh. 42 - Prob. 89CPCh. 42 - Prob. 90CPCh. 42 - Prob. 91CP
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