Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 42, Problem 89CP

(a)

To determine

The allowed distance between positron and electron.

(a)

Expert Solution
Check Mark

Answer to Problem 89CP

The allowed distance between positron and electron is 0.106n2_.

Explanation of Solution

Positron and electron are moving in a circle of radius r2 with opposite velocities. Since the mass and velocity of both the particles are same, their angular momentum is also same.

Write the expression for the total angular momentum of the system.

    Ln=2×mvr2=mvr                                                                                                        (I)

Here, Ln is the total angular momentum of the system, m is the mass of particles, v is the speed, and r is the distance between electron and positron.

Write the expression for the quantized angular momentum.

    Ln=n                                                                                                                    (II)

Here, n is integer, and is the reduced Planck’s constant.

Equate expressions (I) and (II) and solve for v.

    mvr=nv=nmr                                                                                                               (III)

Write the expression for the sum of force acting on the particles.

    F=ma                                                                                                                (IV)

Write the expression for the electrostatic force acting between electron and positron.

    F=kee2r2                                                                                                              (V)

Here, ke is the Coulomb’s constant, e is the charge of electron.

Write the expression for the centripetal acceleration of the particle while moving along a circular path of radius of r2.

    a=v2r2                                                                                                                    (VI)

Use expressions (VI), (V) and (III) in expression (IV) and solve to find r.

    kee2r2=2mv2rkee2r=2mv2=2mn22m2r2r=2n22mkee2                                                                                                     (VII)

Here, n, , m, ke, and e are constants. Take the product of all these terms to be equal to a0.

    a0=2mkee2                                                                                                          (VIII)

    r=2n22mkee2=2a0n2                                                                                                               (IX)

Conclusion:

Substitute 1.06×1034Js for , 9.1×1031kg for m, 8.99×109Nm2/C2 for ke, and 1.6×1019C for e in equation (VIII) to find a0.

    a0=(1.05×1034Js)2(9.1×1031kg)(8.99×109Nm2/C2)(1.602×1019C)2=0.053nm

Substitute 0.053nm for a0 in equation (VIII) to find r.

    r=2(0.053nm)n2=0.106n2

Therefore, the allowed distance between positron and electron is 0.106n2_.

(b)

To determine

The allowed energies of the system.

(b)

Expert Solution
Check Mark

Answer to Problem 89CP

The allowed energies of the system is En=6.80n2_.

Explanation of Solution

Write the expression for the kinetic energy of the system.

    Kt=Ke+Kp                                                                                                         (X)

Here, Kt is the total kinetic energy, Ke is the kinetic energy of electron, and Kp is the kinetic energy of the positron.

Write the expression for electron.

    Ke=12mv2                                                                                                             (XI)

Since the mass and speed of both the particles are same, their kinetic energies are also equal.

    Kp=12mv2                                                                                                           (XII)

Use expression (XI) and (XII) in (X) to find Kt.

    Kt=12mv2+12mv2=mv2                                                                                              (XIII)

Write the expression for the total energy of the system.

    E=Kt+U                                                                                                         (XIV)

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

Write the expression for the electrostatic potential energy of the system.

    U=kee2r                                                                                                        (XV)

Use expressions (XIII) and (XV) in expression (XIV) to find E.

    E=mv2kee2r                                                                                                (XVI)

Use expression (VII).

    kee2r=2mv2mv2=kee22r                                                                                                    (XVII)

Use expression (XVII) in (XVI) to find E.

    E=kee22rkee2r=kee22r                                                                                                (XVIII)

Use expression (VIII) in (XVIII) to find E.

    E=kee22(2a0n2)=kee24a0n2                                                                                                 (XIX)

Conclusion:

Substitute, 8.99×109Nm2/C2 for ke, 1.6×1019C for e, and 0.052nm for a0 in equation (XIX) to find E.

    E=(8.99×109Nm2/C2)(1.06×1019C)24(0.052nm)n2=6.80n2

Therefore, The allowed energies of the system is E=6.80n2_.

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Chapter 42 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

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