Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Question
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Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution
Check Mark

Answer to Problem 3.3P

  r=2.49×1013m=2.49×1011cm

Explanation of Solution

Given:

A µ-meson has a charge of 4.8×1010sC and a mass 207 times that of the resting electron.

Formula used:

The formula to find the radii of the electronic orbits is:

  r=n2h24πme2Zk0

Calculation:

Here,

n = 1

  h=6.625×1034Jsecmmeson=207×9.11×1031kg=1.89×1028kgZ=1k0=9×109Nm2/C2e=4.8×1010sC=4.8×1010sC×((1C3× 109Sc))=1.6×1019C

Substituting the values in the formula of radii of the electronic orbits,

  r=12×(6.625× 10 34Jsec)24π(1.89×1028kg)×(1.6× 10 19C)2×1×9×109Nm2/C2

  r=2.49×1013m=2.49×1011cm

Conclusion:

The radius is r=2.49×1011cm

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution
Check Mark

Answer to Problem 3.3P

E = 2.8 x 103 eV

Explanation of Solution

Given:

A µ-meson has a charge of 4.8×1010sC and a mass 207 times that of the resting electron.

Formula used:

The total energy in any permissible orbit is:

  E=2π2k02mZ2e4h2×1n2

Calculation:

Here,

  n=1h=6.625×1034Jsecmmeson=207×9.11×1031kg=1.89×1028kgZ=1k0=9×109Nm2/C2e=4.8×1010sC=4.8×1010sC×((1C3× 109sC))=1.6×1019C

Substituting the values in the formula of total energy,

  E=2π2 (9× 10 9 N m 2 / C 2 )2×1.89× 10 28kg×12× (1.6× 10 19 C)4 (6.625× 10 34 Jsec)2×112=4.5×1016J×(eV1.6× 10 19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

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