EBK COLLEGE PHYSICS, VOLUME 1
EBK COLLEGE PHYSICS, VOLUME 1
11th Edition
ISBN: 8220103599986
Author: Vuille
Publisher: Cengage Learning US
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Chapter 29, Problem 8P

(a)

To determine

The orbital radii of C12 and C13 .

(a)

Expert Solution
Check Mark

Answer to Problem 8P

The orbital radii of C12 is 7.89 cm.

The orbital radii of C13 is 8.21 cm.

Explanation of Solution

Section 1:

To determine: The orbital radii of C12 .

Answer: The orbital radii of C12 is 7.89 cm.

Given info: Potential (V) is 1.00×103V . The magnetic field strength (B) is 0.200 T.

From Law of conservation of energy,

ΔKE+ΔPE=0 (1)

  • ΔKE is the kinetic energy.
  • ΔPE is the potential energy.

Formula to calculate the kinetic energy is,

ΔKE=12mv2 (2)

  • m is the mass of the atom.
  • v is the velocity.

Formula to calculate the potential energy is,

ΔPE=eV (3)

  • e is the unit of charge.

Substitute Equations (2) and (3) in (1) and re-arrange to get v.

v=2eVm (4)

Formula to calculate the radii is,

r=mveB (5)

Substitute Equation (5) in (4) to get r.

r=2mVeB2 (6)

Substitute 12 u for m, 1.00×103V for V, 0.200 T for B and 1.6×1019C for e in Equation (6) to get r.

r=2(12u)(1.00×103V)(1.6×1019C)(0.200T)2=2(12u)(1.66×1027kg1u)(1.00×103V)(1.6×1019C)(0.200T)2=7.89cm

The orbital radii of C12 is 7.89 cm.

Section 2:

To determine: The orbital radii of C13 .

Answer: The orbital radii of C13 is 8.21 cm.

Given info: Potential (V) is 1.00×103V . The magnetic field strength (B) is 0.200 T.

From Equation (6) of Section 1,

r=2mVeB2

Substitute 13 u for m, 1.00×103V for V, 0.200 T for B and 1.6×1019C for e in the above equation to get r.

r=2(13u)(1.00×103V)(1.6×1019C)(0.200T)2=2(13u)(1.66×1027kg1u)(1.00×103V)(1.6×1019C)(0.200T)2=8.21cm

The orbital radii of C13 is 8.21 cm.

Conclusion:

The orbital radii of C12 is 7.89 cm.

The orbital radii of C13 is 8.21 cm.

(b)

To determine

The relation between the radii of C12 and C13 .

(b)

Expert Solution
Check Mark

Answer to Problem 8P

The relation between the radii of C12 and C13 is r1r2=m1m2 .

Explanation of Solution

Given info: Potential (V) is 1.00×103V . The magnetic field strength (B) is 0.200 T.

From Equation (4) of Section 1,

r=2mVeB2

In the above equation, mass, potential and magnetic field are the same for the atoms. Therefore, r is proportional to m . Therefore,

r1r2=m1m2 (7)

  • r1 and r2 are the radii of C12 and C13 respectively.
  • m1 and m2 are the masses of C12 and C13 respectively.

Substitute 12 u for m1 and 13 u for m2 in the above equation.

r1r2=12u13u=0.961 (8)

From (a), r1=7.89cm and r2=8.21cm . The ratio of r1 and r2 is,

r1r2=7.89cm8.21cm=0.961 (9)

From Equations (8) and (9), radii of C12 and C13 satisfy the formula in (7).

Conclusion:

The relation between the radii of C12 and C13 is r1r2=m1m2

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Chapter 29 Solutions

EBK COLLEGE PHYSICS, VOLUME 1

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