Respuesta :
Answer:
The final size is approximately equal to the initial size due to a very small relative increase of [tex]1.055\times 10^{- 7}[/tex] in its size
Solution:
As per the question:
The energy of the proton beam, E = 250 GeV =[tex]250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J[/tex]
Distance covered by photon, d = 1 km = 1000 m
Mass of proton, [tex]m_{p} = 1.67\times 10^{- 27} kg[/tex]
The initial size of the wave packet, [tex]\Delta t_{o} = 1 mm = 1\times 10^{- 3} m[/tex]
Now,
This is relativistic in nature
The rest mass energy associated with the proton is given by:
[tex]E = m_{p}c^{2}[/tex]
[tex]E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J[/tex]
This energy of proton is [tex]\simeq 250 GeV[/tex]
Thus the speed of the proton, v[tex]\simeq c[/tex]
Now, the time taken to cover 1 km = 1000 m of the distance:
T = [tex]\frac{1000}{v}[/tex]
T = [tex]\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s[/tex]
Now, in accordance to the dispersion factor;
[tex]\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}[/tex]
[tex]\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}[/tex]
Thus the increase in wave packet's width is relatively quite small.
Hence, we can say that:
[tex]\Delta t_{o} = \Delta t[/tex]
where
[tex]\Delta t[/tex] = final width
