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?

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