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

The final size is approximately equal to the initial size due to a very small relative increase of
`1.055* 10^(- 7)` in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =
`250* 10^(9)* 1.6* 10^(- 19) = 4* 10^(- 8) J`

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

Mass of proton,
`m_(p) = 1.67* 10^(- 27) kg`

The initial size of the wave packet,
`\Delta t_(o) = 1 mm = 1* 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* 10^(- 27)* (3* 10^(8))^(2) = 1.503* 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 =
`(1000)/(v)`

T =
`(1000)/(c) = (1000)/(3* 10^(8)) = 3.34* 10^(- 6) s`

Now, in accordance to the dispersion factor;

`(\delta t_(o))/(\Delta t_(o)) = (ht_(o))/(2\pi m_(p)\Delta t_(o)^(2))`

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