Question #d68af

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Question #d68af

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Answer 1 · 2016-05-25T06:18:26

$\approx 1.980 \times 10^{- 11} m$ $approx1.980xx10^-11m$

Explanation:

Heisenberg's uncertainty principle or simply called Uncertainty Principle states that for any particle it is impossible to know simultaneously both position and momentum precisely.
Mathematically stated

$Deltax*Delta p approxh$ .......(1)

where $Deltax$ is uncertainty in position, $Delta p$ is uncertainty in momentum, and $h$ is Planck's Constant and is $6.626xx 10^(-34) m^2 kg // s$ .

Given in the problem uncertainty in speed of a proton $=+-0.01xx10^4ms^-1$
$=> Delta v=0.02xx10^4ms^-1$
$:.$ uncertainty in momentum $Delta p=m_pxxDeltav$
$=1.673 xx 10^-27xx0.02xx10^4$
$=3.346xx10^-25$
Inserting in (1) the known and calulated values
$Deltax approx(6.626xx 10^(-34))/(3.346xx10^-25)$
$approx1.980xx10^-11m$

It is worth noting here that this value is larger than the diameter of the hydrogen nucleus. Which is size of proton itself and is in the range of $1.75 fm (1.75xx10^(−15) m)$ .

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