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^-14min 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 $3.2 \times 10^{- 18} m$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Kinetic energy is, $600 TeV$ .

Step 2: Concept of rest energy

A lead nucleus has mass = $207.2 u$

Thus, the rest energy can be calculated as,

$m c^{2} = (207.2 u) c^{2}$

Step 3:Determine the rest energy

A lead nucleus has mass = $207.2 u$

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

$\begin{array}{rcl} m c^{2} & = & (207.2 u) c^{2} \\ = & (207.2 u c^{2}) (\frac{931.5 MeV}{u c^{2}}) \\ = & 0.193 \times 10^{6} MeV \\ = & 0.193 \times 10^{6} MeV (\frac{10^{6} eV}{1 MeV}) \end{array}$

Convert $MeV$ to $TeV$ as:

$\begin{array}{rcl} m c^{2} & = & 0.193 \times 10^{12} eV (\frac{1 TeV}{10^{12} MeV}) \\ = & 0.193 TeV \end{array}$

Step 4: Determine the kinetic energy lead nucleus

The kinetic energy lead nucleus can be expressed as,

$KE = (γ - 1) m c^{2}$

Here, Y is the relativistic constant.

Re-arrange the equation for Y as,

$γ = \frac{KE}{m c^{2}} + 1$

Substitute $0.193 TeV$ for $m c^{2}$ and $600 TeV$ for $K E$ , and we get,

$\begin{array}{rcl} γ & = & \frac{600 TeV}{0.193 TeV} + 1 \\ = & 3.1 \times 10^{3} \end{array}$

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

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

$l = \frac{l^{'}}{γ}$

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

Substitute $10^{- 14} m$ l for and $3.1 \times 10^{3}$ for Y as shown below.

$\begin{array}{rcl} l^{'} & = & \frac{10^{- 14} m}{3.1 \times 10^{3}} \\ = & 3.2 \times 10^{- 18} m \end{array}$

The thickness of the lead nucleus in the moving direction is $3.2 \times 10^{- 18} m$ .

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Modern Physics

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Short Answer

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

Step by Step Solution

Step 1: Given data

Step 2: Concept of rest energy

Step 3:Determine the rest energy

Step 4: Determine the kinetic energy lead nucleus

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

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Modern Physics

Answers without the blur.

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?

Step by Step Solution

Step 1: Given data

Step 2: Concept of rest energy

Step 3:Determine the rest energy

Step 4: Determine the kinetic energy lead nucleus

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

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