Modern Physics for Scientists and Engineers
Modern Physics for Scientists and Engineers
4th Edition
ISBN: 9781133103721
Author: Stephen T. Thornton, Andrew Rex
Publisher: Cengage Learning
Question
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Chapter 4, Problem 56P

(a)

To determine

The energies of the three lowest states of positronium.

(a)

Expert Solution
Check Mark

Answer to Problem 56P

The energies of the three lowest states of positronium is 6.803eV, 1.701eV and 0.756eV.

Explanation of Solution

Positronium is an atom composed of an electron and a Positron and the mass of positron is equal to the mass of electron.

Write the expression to calculate reduced mass.

μ=mMm+M        (I)

Here, μ is the reduced mass of an atom, m is the mass of positron and M is the mass of electron.

Since the mass of positron is equal to the mass of electron above equation can be written as:

    μ=mmm+m=m2

Write the expression for effective Rydberg constant.

    Reff=R2

Here, Reff is the effective Rydberg constant and R is the Rydberg constant for infinite nuclear mass.

Write the expression for energy at nth level.

    En=hc(Reff)n2        (II)

Here, En is the energy at nth level, h is the Planck’s constant, and c is the speed of light.

Conclusion:

Substitute 1239.8eVnm for hc , 1.097373×102/2nm-1 for Reff and 1 for n in equation (II).

    E1=(1239.8eVnm)(1.097373×1022nm-1)12=6.803eV

Similarly for n=2 energy is E2=1.701eV, for n=3 energy is E3=0.756eV and for n=4 energy is E4=0.425eV

Thus, the energies of the three lowest states of positronium is 6.803eV, 1.701eV and 0.756eV.

(b)

To determine

The wavelengths of the Kα,Kβ,LαandLβ transitions.

(b)

Expert Solution
Check Mark

Answer to Problem 56P

The wavelengths of the Kα transition is 243.0nm ,Kβ transitions is 205.0nm ,Lα transition is 1312nm and Lβ transition is 971.6nm.

Explanation of Solution

Write the expression for wavelength.

    λ=hcEuEl        (III)

Here, λ is the wavelength of transition, Eu is energy of upper energy state and El is the energy of lower energy state.

Conclusion:

For λ(Kα) : nl=1andnu=2.

Substitute 1239.8eVnm for hc, 1.701eV for Eu and 6.803eV for El in equation (III).

    λ=1239.8eVnm1.701eV(6.803eV)=243.0nm

For λ(Kβ) : nl=1andnu=3.

Substitute 1239.8eVnm for hc, 0.756eV for Eu and 6.803eV for El in equation (III).

    λ=1239.8eVnm0.756eV(6.803eV)=205.0nm

For λ(Lα) : nl=2andnu=3.

Substitute 1239.8eVnm for hc, 0.756eV for Eu and 1.701eV for El in equation (III).

    λ=1239.8eVnm0.756eV(1.701eV)=1312nm

For λ(Lβ) : nl=2andnu=4.

Substitute 1239.8eVnm for hc, 0.425eV for Eu and 1.701eV for El in equation (III).

    λ=1239.8eVnm0.425eV(1.701eV)=971.6nm

Thus, the wavelengths of the Kα transition is 243.0nm ,Kβ transitions is 205.0nm ,Lα transition is 1312nm and Lβ transition is 971.6nm.

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