PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 19, Problem 128P

(a)

To determine

The speed with which the 238U2+ ions emerge from the accelerating plates.

(a)

Expert Solution
Check Mark

Answer to Problem 128P

The speed with which the 238U2+ ions emerge from the accelerating plates is 2v .

Explanation of Solution

The kinetic energy gained by the ion must be equal to the potential energy lost.

Write the equation for the conservation of energy.

  ΔK=ΔU        (I)

Here, ΔK is the gain kinetic energy of the ion and ΔU is the loss in potential energy by the ion.

Write the equation for ΔK .

  ΔK=12mv2        (II)

Here, m is the mass of the ion.

Write the equation for ΔU .

  ΔU=qΔV        (III)

Here, q is the charge of the ion and ΔV is the potential difference through which the ion is accelerated.

Put equations (II) and (III) in equation (I).

  12mv2=qΔVv=2qΔVmvqv=kq        (IV)

Here, k is the proportionality constant.

Write the relation for the speed of 235U+ ions using relation (IV).

  v2+=kq2+

Here, v2+ is the speed of the 238U2+ ions and q2+ is the charge of the 238U2+ ions.

Write the relation for the speed of 238U+ ions using relation (IV).

  v+=kq+

Here, v+ is the speed of the 238U+ ions and q+ is the charge of the 238U+ ions.

Take the ratio of the above two equations.

  v2+v+=kq2+kq+=q2+q+v2+=q2+q+v+        (V)

Conclusion:

The charge of the ion 238U+ ions is equal to +e and that of 238U2+ ions is +2e where e is the magnitude of electronic charge.

Substitute +2e for q2+ , +e for q+ and v for v+ in equation (V) to find v2+ .

  v2+=+2e+ev=2v

Therefore, the speed with which the 238U2+ ions emerge from the accelerating plates is 2v .

(b)

To determine

The sketch of the trajectory of 238U2+ ions inside the velocity selector.

(b)

Expert Solution
Check Mark

Answer to Problem 128P

The sketch of the trajectory of 238U2+ ions inside the velocity selector is

PHYSICS, Chapter 19, Problem 128P , additional homework tip  1

Explanation of Solution

In a velocity selector, for an ion with speed v , the electric and magnetic fields will be equal in magnitude and opposite in direction. For an ion with speed greater than v , the magnetic force will dominate. In part (a), it is found that the speed of the 238U2+ ions is 2v which is greater than v . This implies for 238U2+ ions magnetic force dominates.

Write the equation for the magnetic force.

  FB=qv×B        (VI)

Here, FB is the magnetic force, v is the velocity of the charge and B is the magnetic field.

In the velocity selector, FE must oppose FB . The direction of the magnetic force will be perpendicular to both velocity vector and magnetic field according to equation. In accordance with the right-hand rule, FB will be directed downward for a positive ion. The sketch of the 238U2+ ions inside the velocity selector is shown below.

PHYSICS, Chapter 19, Problem 128P , additional homework tip  2

Conclusion:

Thus, the sketch of the 238U2+ ions inside the velocity selector is sketched in figure 1.

(c)

To determine

The diameter of the path of 238U2+ ions.

(c)

Expert Solution
Check Mark

Answer to Problem 128P

The diameter of the path of 238U2+ ions is D2 .

Explanation of Solution

Write the equation for the magnitude of the magnetic force on ions using equation (V).

  FB=qvB        (VI)

Here, FB is the magnitude of the magnetic force and B is the magnitude of the magnetic field.

The magnetic force provides the centripetal force needed for the circular motion of the ions.

Write the relationship between the magnetic force and the centripetal force.

  FB=FC        (VII)

Here, FC is the magnitude of the centripetal force.

Write the equation for the magnitude of the centripetal force on ions.

  FC=mv2r        (VIII)

Here, r is the radius of the circular path.

Write the equation for the radius of a circle.

  r=D2

Here, D is the diameter of the circular path.

Put the above equation in equation (VIII).

  FC=mv2(D2)=m2v2D        (IX)

Put equations (VI) and (IX) in equation (VII).

  qvB=m2v2DB2m=vqDconstant=vqD

Since m and B are constants for the two kinds of ions, the left-hand side of the above equation must be constant. This implies the right-hand side of the above equation is also constant.

Write the relationship between the diameters of the paths of 238U+ ions and 238U2+ ions using the above relation.

  vqD=v2+q2+D2+

Here, v is the speed of the 238U+ ions, D is the diameter of the circular path of 238U+ ions , q is the charge of the 238U+ ions and D2+ is the diameter of the circular path of 238U2+ ions.

Rewrite the above relationship for D2+ .

  D2+=v2+qvq2+D        (X)

Conclusion:

Substitute 2v for v2+ , +e for q and +2e for q2+ in equation (X) to find D2+ .

  D2+=(2v)(+e)v(+2e)D=22D=D2

Therefore, the diameter of the path of 238U2+ ions is D2 .

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Chapter 19 Solutions

PHYSICS

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