# The velocity of a particle is 5.8*10^5 "m/s" . Calculate the uncertainty in its position. (Mass of particle=9.1x10^(-28)"g" , h=6.63*10^(-34))?

Jul 11, 2015

You can use Heisenberg's Uncertainty Principle:

#### Explanation:

Assuming the particle moving along the x-axis we can use Heisenberg's Uncertainty Principle:
$\Delta x \Delta {p}_{x} \ge \frac{h}{4 \pi}$
where:
$x$ is position
${p}_{x}$ is momentum;
$h$ is Planck's Constant;
and the $\Delta$s represent uncertainties.
With your data you can consider:

1] a perfect knowledge of the velocity, implying an uncertainty of zero in the momentum (assuming a constant mass): $\Delta {p}_{x} = 0$
This gives you an enormous value (tending towards $\infty$) of the uncertainty in the position (you do not know where the particle is!!!);

2] The uncertainty in velocity will be $\pm 5.8 \times {10}^{5} \frac{m}{s}$ meaning that the velocity you have is only one of the velocities you measured (the highest, probably, with other measurement representing possible velocities in a certain range from $0$ to $5.8 \times {10}^{5} \frac{m}{s}$).
So $\Delta {p}_{x} = \frac{9.1 \times {10}^{-} 28}{1000} \cdot 5.8 \times {10}^{5} = 5.28 \times {10}^{-} 25 k g \frac{m}{s}$
and:
$\Delta x \ge \frac{h}{\Delta {p}_{x} 4 \pi} = 9.99 \times {10}^{-} 11 m$
considering that the radius of an atom is $\approx {10}^{-} 10 m$!!!

3] I thought of an intriguing situation!!! Observing your velocity, I noticed that is quite high!!!
So, I thought to introduce an uncertainty in the momentum given by the difference between:
Classical momentum : ${p}_{x} = m v = \frac{9.1 \times {10}^{-} 28}{1000} \cdot 5.8 \times {10}^{5} = 5.28 \times {10}^{-} 25 k g \frac{m}{s}$
Relativistic momentum :
${p}_{x r} = \frac{m v}{\sqrt{1 - {v}^{2} / {c}^{2}}} = 5.31 \times {10}^{-} 25 k g \frac{m}{s}$
(where: $c = 3 \times {10}^{8} \frac{m}{s}$ is the speed of light)
So $\Delta {p}_{x} =$3xx10^-27kgm/s#

$\Delta x \ge \frac{h}{\Delta {p}_{x} \cdot 4 \cdot \pi} = 1.76 \times {10}^{-} 8 m$