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Q26E

Expert-verified
Modern Physics
Found in: Page 557
Modern Physics

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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Short Answer

A lead nucleus at rest is roughly 10-14 min diameter. If moving through the laboratory with a kinetic energy of 600 TeV, howthick would thenucleus be in the direction of motion?

The thickness of the lead nucleus in the moving direction is 3.2×10-18 m

See the step by step solution

Step by Step Solution

Step 1: Given data

Kinetic energy is, 600 TeV .

Step 2: Concept of rest energy

A lead nucleus has mass = 207.2u

Thus, the rest energy can be calculated as,

mc2=(207.2u)c2

Step 3:Determine the rest energy

A lead nucleus has mass = 207.2u

Thus, the rest energy can be calculated as shown below:

mc2=(207.2u)c2=207.2uc2931.5 MeVuc2=0.193×106 MeV=0.193×106 MeV106 eV1 MeV

Convert MeV to TeV as:

mc2=0.193×1012 eV1 TeV1012 MeV=0.193 TeV

Step 4: Determine the kinetic energy lead nucleus

The kinetic energy lead nucleus can be expressed as,

KE=(γ-1)mc2

Here, Y is the relativistic constant.

Re-arrange the equation for Y as,

γ=KEmc2+1

Substitute 0.193 TeV for mc2 and 600 TeV for KE , and we get,

γ=600 TeV0.193 TeV+1=3.1×103

Step 5: Determine the thickness of the lead nucleus in the moving direction

Thus, the length in the moving direction can be expressed as:

l=l'γ

Here l is the thickness of the lead nucleus when it rest.

Substitute 10-14 m l for and 3.1×103 for Y as shown below.

l'=10-14 m3.1×103=3.2×10-18 m

The thickness of the lead nucleus in the moving direction is 3.2×10-18 m .

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