The radius of the first Bohr Orbit.

Question

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Answer 1

Question

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

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Answer 2

Question

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

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Answer 3

Question

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

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Answer 4

Question

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

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Answer 5

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

Answer 6

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

Answer 7

Chapter 3, Problem 3.3P

(a)

To determine

To calculate: The radius of the first Bohr Orbit.

Answer 8

(a)

Expert Solution

Answer to Problem 3.3P

r=2.49×10−13m=2.49×10−11cm

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109Sc))=1.6×10−19C

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

r=12×(6.625× 10 −34Jsec)24π(1.89×10−28kg)×(1.6× 10 −19C)2×1×9×109Nm2/C2

r=2.49×10−13m=2.49×10−11cm

Conclusion:

The radius is r=2.49×10−11cm

Answer 13

(b)

To determine

To calculate: The ionization potential.

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

Answer 14

(b)

To determine

To calculate: The ionization potential.

Answer 15

(b)

Expert Solution

Answer to Problem 3.3P

E = 2.8 x 10³ eV

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

A µ-meson has a charge of −4.8×10−10sC 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=6.625×10−34Jsecmmeson=207×9.11×10−31kg=1.89×10−28kgZ=1k0=9×109Nm2/C2e=4.8×10−10sC=4.8×10−10sC×((1C3× 109sC))=1.6×10−19C

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=4.5×10−16J×(eV1.6× 10 −19C)=2.8×103eV

Conclusion:

The ionization potential =2.8×103eV

Answer 20

Ch. 3 - Prob. 3.1P Ch. 3 - Prob. 3.2P Ch. 3 - Prob. 3.3P Ch. 3 - Prob. 3.4P Ch. 3 - Calculate the current due to the hydrogen electron...Ch. 3 - Prob. 3.6P Ch. 3 - Prob. 3.7P Ch. 3 - Prob. 3.8P Ch. 3 - Prob. 3.9P Ch. 3 - Prob. 3.10P

The radius of the first Bohr Orbit.

Concept explainers

To calculate: The radius of the first Bohr Orbit.

Answer to Problem 3.3P

Explanation of Solution

To calculate: The ionization potential.

Answer to Problem 3.3P

Explanation of Solution

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Chapter 3 Solutions