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Chapter 29, Problem 10P

(a)

To determine

The uncertainty in momentum of electron in terms of r.

(a)

Expert Solution
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Answer to Problem 10P

The uncertainty in momentum is Δp  2r.

Explanation of Solution

Write the equation for uncertainty principle.

    ΔxΔp  2

Here, Δx is the uncertainty in position, Δp is the uncertainty in momentum and is the reduced Plank’s constant.

Conclusion:

Rewrite the above relation by substituting r for x.

    (r)Δp  2

Rewrite the above relation in terms of Δp.

    Δp  2(r)

Therefore, the uncertainty in momentum is Δp  2r.

(b)

To determine

The kinetic energy of electron.

(b)

Expert Solution
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Answer to Problem 10P

The kinetic energy of electron will be 22mer2.

Explanation of Solution

Write the equation for momentum.

    K=(Δp)22me

Here, K is the kinetic energy and me is the mass of electron.

Conclusion:

Rewrite the above equation by substituting  2r for Δp.

  K=( 2r)22me=22mer2

Therefore, the kinetic energy of electron will be 22mer2.

(c)

To determine

The total energy of electron in terms of r.

(c)

Expert Solution
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Answer to Problem 10P

The total energy of electron in terms of r is 22mer2kee2r.

Explanation of Solution

Write the equation for total energy of electron.

    E=K+U

Here, E is the total energy and U is the potential energy.

Write the equation for U.

  U=kee2r

Here, k is the electrostatic constant and e is the magnitude of electric charge.

Rewrite the equation for E by substituting the above relation for U and 22mer2 for K.

Conclusion:

Rewrite the above equation by substituting  2r for Δp.

  E=22mer2+(kee2r)=22mer2kee2r                                                                                   (I)

Therefore, the total energy of electron in terms of r is 22mer2kee2r.

(d)

To determine

The value of r.

(d)

Expert Solution
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Answer to Problem 10P

The value of r is 2mekee2.

Explanation of Solution

Write the condition for the value of E to be minimum.

    dEdr=0

Conclusion:

Rewrite the above equation by substituting for 22mer2kee2r for E.

  dEdr=0d(22mer2kee2r)dr=02mer3+kee2r2=0

Rewrite the above relation in terms of r.

  r=2mekee2

The calculated value 2mekee2 is the Bohr radius.

Therefore, the value of r is 2mekee2.

(e)

To determine

The resulting total energy.

(e)

Expert Solution
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Answer to Problem 10P

The resulting total energy is 13.6eV.

Explanation of Solution

Write the equation for E found in part (c).

    E=22mer2kee2r

Rewrite the above equation by substituting 2mekee2 for r.

    E=22me(2mekee2)2kee2(2mekee2)=meke2e422

Conclusion:

Substitute 9.109×1031kg for me, 8.988×109 Nm2/C2 for ke, 1.602×1019 C for e, and 1.055×1034 Js for in the above equation to find E.

    E=(9.109×1031kg)(8.988×109Nm2/C2)2(1.602×1019C)42(1.055×1034Js)2=2.179×1018 J(1 eV1.602×1019 J)=13.6eV

Therefore, the resulting total energy is 13.6eV.

(f)

To determine

States that the answers are in agreement the conclusions of Bohr theory.

(f)

Expert Solution
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Explanation of Solution

The uncertainty of momentum of electron calculated in part (a) is Δp  2r. It is two times the minimum possible value of momentum. This makes the result is in agreement with the conclusions of Bohr theory.

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Chapter 29 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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