EBK AN INTRODUCTION TO MODERN ASTROPHYS
EBK AN INTRODUCTION TO MODERN ASTROPHYS
2nd Edition
ISBN: 9781108390248
Author: Carroll
Publisher: YUZU
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Chapter 5, Problem 5.8P

(a)

To determine

The expression for orbital radii and energies.

(a)

Expert Solution
Check Mark

Answer to Problem 5.8P

The expression for orbital radii is 4πε0n2h2μze2 and the expression for orbital energy is μ2z2e432π2ε02n2h2.

Explanation of Solution

Write the expression for force between two charges.

    F=14πε0q1q2r2        (1)

Here, q1 and q2 are the electric charges, r is the distance between the charges are z is the number of proton in the atom.

Write the expression for electric potential energy between two charges.

    U=14πε0q1q2r        (2)

Write the expression for kinetic energy of the electron.

    K=12μv2        (3)

Here, μ is the mass of the electron and v is the velocity of the electron.

Write the expression for centripetal force between electron and the nucleus.

    Fc=μv2r        (4)

Write the expression for Bohr’s quantization of angular momentum.

    μvr=nh        (5)

Here, n is the number of orbit.

Write the expression for total orbital energy.

    E=U+K        (6)

Conclusion:

Substitute e for q1, ze for q2 in the equation (1).

    F=14πε0ze2r2        (7)

The centripetal force is equal to force between two charges.

Substitute μv2r for F in the equation (7)

    μv2r=14πε0ze2r2μv2=14πε0ze2r12μv2=18πε0ze2r        (8)

Substitute nhμr for v in the equation (8).

    12μ(nhμr)2=18πε0ze2r12n2h2μr2=18πε0ze2rr=4πε0n2h2μze2=0.0529n2Z

Substitute 12μv2 for K and 14πε0q1q2r for U in the Equation (6).

    E=12μv214πε0q1q2r        (9)

Substitute nhμr for v, 4πε0n2h2μze2 for r in the equation (9).

    E=12μ(nhμr)214πε0q1q2(4πε0n2h2μze2)=12n2h2μ(4πε0n2h2μze2)214πε0q1q2(4πε0n2h2μze2)=μ2z2e432π2ε02n2h2

Thus, the expression for orbital radii is 4πε0n2h2μze2 and the expression for orbital energy is μ2z2e432π2ε02n2h2.

(b)

To determine

The radius and energy of ground state orbit and ionization energy of the singly ionized helium.

(b)

Expert Solution
Check Mark

Answer to Problem 5.8P

The radius of ground state orbit is 0.0265nm and the ionization energy of the singly ionized helium is 54.4eV.

Explanation of Solution

Write the expression for radius of ground state orbit.

    r=0.0529n2Z        (10)

Write the expression for energy of ground state orbit.

    E=(13.6eV)Z2n2        (11)

Conclusion:

Substitute 2 for Z and 1 for n in the equation (10).

    r=0.0529nm(1)2(2)=0.0265nm

Substitute 2 for Z and 1 for n in the equation (11).

    E=(13.6eV)(2)2(1)2=54.4eV

Thus, the radius of ground state orbit is 0.0265nm and the ionization energy of the singly ionized helium is 54.4eV.

(c)

To determine

The radius and energy of ground state orbit or ionization energy of the doubly ionized lithium.

(c)

Expert Solution
Check Mark

Answer to Problem 5.8P

The radius of ground state orbit is 0.0176nm and the ionization energy of the doubly ionized lithium is 122.4eV.

Explanation of Solution

The equation (10) and the equation (11) is expression for ground state radius and the ionization energy.

Conclusion:

Substitute 3 for Z and 1 for n in the equation (10).

    r=0.0529nm(1)2(3)=0.0176nm

Substitute 3 for Z and 1 for n in the equation (11).

    E=(13.6eV)(3)2(1)2=122.4eV

Thus, the radius of ground state orbit is 0.0176nm and the ionization energy of the doubly ionized lithium is 122.4eV.

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