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Q26E

Expert-verifiedFound in: Page 557

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 $3.2\times {10}^{-18}\text{m}$

** **

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

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

**Thus, the rest energy can be calculated as,**

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

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

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

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

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

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

The kinetic energy lead nucleus can be expressed as,

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

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,

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

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 $3.1\times {10}^{3}$ for Y as shown below.

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

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

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