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Q26E

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Found in: Page 557

### Modern Physics

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

# 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 $3.2×{10}^{-18}\text{m}$

See the step by step solution

## Step 1: Given data

Kinetic energy is, $600\text{TeV}$ .

## Step 2: Concept of rest energy

A lead nucleus has mass = ${207}{.}{2}{\text{u}}$

Thus, the rest energy can be calculated as,

${m}{{c}}^{{2}}{=}{\left(}{207}{.}{2}{\text{u}}{\right)}{{c}}^{{2}}$

## Step 3:Determine the rest energy

A lead nucleus has mass = $207.2\text{u}$

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

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

Convert $\text{MeV}$ to $\text{TeV}$ as:

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

## Step 4: Determine the kinetic energy lead nucleus

The kinetic energy lead nucleus can be expressed as,

$\text{KE}=\left(\gamma -1\right)m{c}^{2}$

Here, Y is the relativistic constant.

Re-arrange the equation for Y as,

$\gamma =\frac{\text{KE}}{m{c}^{2}}+1$

Substitute $0.193\text{TeV}$ for $m{c}^{2}$ and $600\text{TeV}$ for $KE$ , and we get,

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

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

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

$l=\frac{{l}^{\text{'}}}{\gamma }$

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

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

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

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