Question

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Answer 1

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =$250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v$\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width