Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 42, Problem 80AP

(a)

To determine

Energy of electron in ground state and the first four excited state.

(a)

Expert Solution
Check Mark

Answer to Problem 80AP

Energy of the electron in ground state is 8.16eV_, electron in first excited state is 2.04eV_, electron in second excited state is 0.902eV_, electron in third excited state is 0.508eV_, and electron in fourth excited state is 0.325eV_.

Explanation of Solution

Write the expression for the ground state energy of atom.

    E1=hcλseries limit                                                                                                (I)

Here, E1 is the ground state energy of atom, h is Planck’s constant, c is the speed of light, and λseries limit is the wavelength.

Write the expression for difference in energy between first and ground state energy level.

    ΔE21=hcλ1                                                                                                        (II)

Here, ΔE21 is the energy difference between first and ground state energy level., and λ1 is the wavelength of the light emitted during transition from first to ground state.

Write the expression to find E2.

    E2=E1+ΔE21                                                                                                        (III)

Here, E2 is the energy of first excited state.

Write the expression for the difference in energy between second excited level and ground state.

    ΔE31=hcλ2                                                                                                          (IV)

Here, ΔE31 is the energy difference between second and ground state energy level., and λ2 is the wavelength of the light emitted during transition from second to ground state.

Write the expression to find energy of second excited level.

    E3=E2+ΔE31                                                                                                      (V)

Here, E3 is the energy of second excited state.

Write the expression for the difference in energy between third excited level and ground state.

    ΔE41=hcλ3                                                                                                          (VI)

Here, ΔE41 is the energy difference between third and ground state energy level., and λ3 is the wavelength of the light emitted during transition from third to ground state.

Write the expression to find energy of third excited level.

    E4=E3+ΔE41                                                                                                      (VII)

Here, E4 is the energy of third excited state.

Write the expression for the difference in energy between fourth excited level and ground state.

    ΔE51=hcλ4                                                                                                        (VIII)

Here, ΔE51 is the energy difference between fourth and ground state energy level., and λ4 is the wavelength of the light emitted during transition from fourth to ground state.

Write the expression to find energy of fourth excited level.

    E5=E4+ΔE51                                                                                                        (IX)

Here, E5 is the energy of fourth excited state.

Conclusion:

Substitute 1240eVnm for hc, and 152.0nm for λseries limit in expression (I) to find E1.

    E1=1240eVnm152.0nm=8.16eV

Substitute 1240eVnm for hc, and 202.6nm for λ1 in equation (II) to find ΔE21.

    ΔE21=1240eVnm(202.6nm)=6.12eV

Substitute 8.16eV for E1, and 6.12eV for ΔE21 in equation (III) to find E2.

    E2=8.16eV+6.12eV=2.04eV

Substitute 1240eVnm for hc, and 170.9nm for λ2 in equation (IV) to find ΔE31.

    ΔE31=1240eVnm(170.9nm)=7.26eV

Substitute 8.16eV for E1, and 7.26eV for ΔE31 equation (V)to find E3.

    E3=8.16eV+7.26eV=0.902eV

Substitute 1240eVnm for hc, and 162.1nm for λ3 in equation (VI) to find ΔE41.

    ΔE41=1240eVnm(162.1nm)=7.65eV

Substitute 8.16eV for E1, and 7.65eV for ΔE41 equation (VII)to find E4.

    E4=8.16eV+7.65eV=0.508eV

Substitute 1240eVnm for hc, and 158.3nm for λ4 in equation (VIII) to find ΔE51.

    ΔE51=1240eVnm(158.3nm)=7.83eV

Substitute 8.16eV for E1, and 7.83eV for ΔE51 equation (IX)to find E5.

    E5=8.16eV+7.83eV=0.325eV

Therefore, the energy of the electron in ground state is 8.16eV_, electron in first excited state is 2.04eV_, electron in second excited state is 0.902eV_, electron in third excited state is 0.508eV_, and electron in fourth excited state is 0.325eV_.

(b)

To determine

The wavelength of the first three lines and the short wavelength limit in the Balmer series.

(b)

Expert Solution
Check Mark

Answer to Problem 80AP

The wavelength of the α line is 1090nm_, β line is 809nm_, γ line is 724nm_, and the short wavelength limit is 609nm_.

Explanation of Solution

In Balmer series transition takes place between ith level and second level.

Write the expression for the difference in energy between the energy levels for Balmer series.

    EiE2=hcλ                                                                                                   (X)

Here, Ei is the energy corresponding to the ith energy level.

Rearrange expression (X) to find λ.

    λ=hcEiE2                                                                                                           (XI)

Put i=3 in equation (XI) to find expression for wavelength of α line.

    λ3=hcE3E2                                                                                                      (XII)

Put i=4 in equation (XI) to find expression for wavelength of β line.

    λ4=hcE4E2                                                                                                      (XIII)

Put i=5 in equation (XI) to find expression for wavelength of γ line.

    λ5=hcE5E2                                                                                                      (XIV)

Put i= in equation (XI) to find expression for wavelength of shortest wave.

    λ=hcEE2                                                                                                      (XV)

Conclusion:

Substitute 0.0902eV for E3, 1240nmeV for hc, and 2.04eV for E2 in expression (XII) to find λ3.

    λ3=1240nmeV(0.902eV)(2.04eV)=1090nm

Substitute 0.508eV for E4, 1240nmeV for hc, and 2.04eV for E2 in expression (XIII) to find λ4.

    λ4=1240nmeV(0.508eV)(2.04eV)=809nm

Substitute 0.325eV for E5, 1240nmeV for hc, and 2.04eV for E2 in expression (XIV) to find λ5.

    λ5=1240nmeV(0.325eV)(2.04eV)=724nm

Substitute for E, 1240nmeV for hc, and 2.04eV for E2 in expression (XV) to find λ.

    λ=1240nmeV()(2.04eV)=609nm

Therefore, The wavelength of the α line is 1090nm_, β line is 809nm_, that of γ line is 724nm_, and that of short wavelength is 609nm_.

(c)

To determine

Show that the first four lines and the shortest wavelengths of Lyman series is equal to 60% of the wavelength for Lyman series of a one electron system.

(c)

Expert Solution
Check Mark

Answer to Problem 80AP

It is shown that the first four lines and the shortest wavelengths of Lyman series is equal to 60% of the wavelength for Lyman series of a one electron system.

Explanation of Solution

Write the expression for the wavelength of Lyman series.

    1λ=RH(11n2)                                                                                          (XVI)

Here, RH is the Rydberg constant, and n is the principle quantum number.

Rearrange expression (XVI) to find λ.

    λ=1RH(11n2)                                                                                             (XVII)

Conclusion:

For α series n=2. Put 2 for n, 10973731.6m1 for RH in expression (XVI) to find wavelength of α line.

    λα=110973731.6m1(1122)=122nm

Put 3 for n, 10973731.6m1 for RH in expression (XVII) to find wavelength of β line.

    λβ=110973731.6m1(1132)=103nm

Put 4 for n, 10973731.6m1 for RH in expression (XVII) to find wavelength of γ line.

    λγ=110973731.6m1(1142)=97.2nm

Put 5 for n, 10973731.6m1 for RH in expression (XVII) to find wavelength of δ line.

    λδ=110973731.6m1(1152)=94.9nm

Put for n, 10973731.6m1 for RH in expression (XVII) to find wavelength of short wavelength.

    λ=110973731.6m1(112)=91.1nm

60% of the first spectral line of wavelength 202.6nm is 0.600(202.6nm)=122nm.

60% of the second spectral line of wavelength 170.9nm is 0.600(170.9nm)=103nm.

60% of the third spectral line of wavelength 162.1nm is 0.600(162.1nm)=97.3nm.

60% of the fourth spectral line of wavelength 158.3nm is 0.600(158.3nm)=95.0nm.

60% of the fifth spectral line of wavelength 2152.0nm is 0.600(150.0nm)=91.2nm.

Comparing the 60% value of wavelengths of spectral lines of the energy diagram in question, it is seen that the values are equal to the wavelength of α,β,γ,δ, and the short wavelength.

Therefore, it is shown that the first four lines and the shortest wavelengths of Lyman series of Hydrogen atom is equal to 60% of the wavelength for Lyman series of a one electron system.

(d)

To determine

The reason why the atom can be assumed to be Hydrogen.

(d)

Expert Solution
Check Mark

Answer to Problem 80AP

The atom can be Hydrogen since the results of Doppler shifts is equal to the actual value of wavelength of Hydrogen spectrum.

Explanation of Solution

The wavelength of the spectral lines were measured by the Doppler shifting in the frequency and speed of motion of atom with respect to the Earth.

Write the expression for the ratio of original wavelength and the shifted wavelength of the atom.

    λλ=cvc+v (XVIII)

The ratio between the old wavelengths and Doppler shifted wavelength is equal to 0.600.

    λλ=cvc+v=0.600                                                                                  (XIX)

Square equation (XIX) and solve for v.

    cvc+v=(0.600)2cv=0.36c+0.36v0.64c=1.36vv=0.471c

Thus if we multiply the shifted wavelength by 0.600 we will get wavelengths that correspond to the wavelength of hydrogen spectrum. Here the source is moving with a speed of 0.471c.

Conclusion:

Therefore, the atom can be Hydrogen since the results of Doppler shifts is equal to the actual value of wavelength of Hydrogen spectrum.

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Chapter 42 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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