PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Question
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Chapter 39, Problem 9P

(a)

To determine

The length of each bundle in its own reference frame.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

In laboratory frame of reference length of each bundle is 1.0cm .

Diameter of each bundle in laboratory frame of refence is 10μm .

Each particle has energy of 50GeV .

Formula used:

Write expression for energy of beam.

  E=γmc2

Solve above expression for γ .

  γ=Emc2

Write expression for proper length.

  Lp=γL

Substitute Emc2 for γ in above expression.

  Lp=Emc2L  ....... (1)

Calculation:

Substitute 50Gev for E , 0.511MeV for mc2 and 1.0cm for L in equation (1).

  Lp=50GeV0.511MeV(109eV1GeV)(1MeV106eV)(1.0cm)(1m100cm)Lp=978.5m

Conclusion:

Thus, length of each bundle is 978.5m .

(b)

To determine

The minimum proper length of accelerator.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

In laboratory frame of reference length of each bundle is 1.0cm .

Diameter of each bundle in laboratory frame of refence is 10μm .

Each particle has energy of 50GeV .

Formula used:

Write expression for energy of beam.

  E=γmc2

Solve above expression for γ .

  γ=Emc2

Write expression for proper length.

  Lp=γL

Substitute Emc2 for γ in above expression.

  Lp=Emc2L  ....... (1)

Write expression for proper length of accelerator.

  Lp=Lp,accγ

Solve above expression for Lp.acc .

  Lp,acc=γLp

Substitute Emc2 for γ in above expression.

  Lp,acc=Emc2Lp  ....... (2)

Calculation:

Substitute 50Gev for E , 0.511MeV for mc2 and 1.0cm for L in equation (1).

  Lp=50GeV0.511MeV(109eV1GeV)(1MeV106eV)(1.0cm)(1m100cm)Lp=978.5m

Substitute 50Gev for E , 0.511MeV for mc2 and 978.5m for Lp in equation (2).

  Lp,acc=50GeV0.511MeV(109eV1GeV)(1MeV106eV)(978.5m)Lp,acc=9.6×107m

Conclusion:

Thus, the minimum proper length of accelerator is 9.6×107m .

(c)

To determine

The length of positron bundle which moves in the refence of electron bundle.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

In laboratory frame of reference length of each bundle is 1.0cm .

Diameter of each bundle in laboratory frame of refence is 10μm .

Each particle has energy of 50GeV .

Formula used:

Write expression for energy of beam.

  E=γmc2

Solve above expression for γ .

  γ=Emc2

Write expression for length of positron bundle.

  Lpos=Lγ

Substitute Emc2 for γ in above expression.

  Lpos=LEmc2  ....... (1)

Calculation:

Substitute 50Gev for E , 0.511MeV for mc2 and 1.0cm for L in equation (1).

  Lpos=1.0cm(1m100cm)50GeV0.511MeV(109eV1GeV)(1MeV106eV)Lpos=0.1×106m(1μm106m)Lpos=0.1μm

Conclusion:

Thus, the length of positron is 0.1μm .

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