# Question aa03b

Feb 15, 2016

We need to consider two things
(A). Heisenberg's uncertainty principle
(B). Escape Velocity of electron to break away from the potential energy of nucleus-electron system.

#### Explanation:

(A). 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

$\Delta x \cdot \Delta p \approx h$ .......(1)

where $\Delta x$ is uncertainty in position, $\Delta p$ is uncertainty in momentum, and $h$ is Planck's Constant and is $6.626 \times {10}^{- 34} {m}^{2} k g / s$.
If an electron is confined to nucleus, uncertainty in position of electron will be size of the nucleus itself. The diameter of the hydrogen nucleus is in the range of 1.75 fm (1.75xx10^(−15) m).
From equation (1), uncertainty in momentum is given by
$\Delta p \approx \frac{h}{\Delta x}$
approx(6.626xx 10^(-34))/(1.75xx10^(−15))

Inserting value of mass of an electron we obtain uncertainty in the velocity of electron as
Delta vapprox(6.626xx 10^(-34))/(1.75xx10^(−15)xx(9.11×10^(-31))#

$= 4.15 \times {10}^{11} m / s$

The uncertainty itself is more than speed of light in vacuum. We know from postulate of Special Theory of relativity that no material particle can travel at speed of light in vacuum. Hence the result is inconsistent.

(B). Electric Potential Energy of an electron in the nucleus is given by the expression
$P E = - {e}^{2} / \left(4 \pi {\epsilon}_{\circ} a\right)$
Where $e$ is electronic charge, $a$ is radius of the nucleus, $4 \pi {\epsilon}_{\circ}$ is constant.
$P E \approx - \left(8.988 \times {10}^{9}\right) {\left(1.6 \times {10}^{-} 19\right)}^{2} / \left({10}^{-} 15\right)$

$P E \approx - 2.3 \times {10}^{-} 13 J$

For the electron to escape from the nucleus to infinity, total energy must be zero. If ${v}_{e}$ is the escape velocity of the electron, its Kinetic energy is given by
$K E = \frac{1}{2} {m}_{e} {v}_{e}^{2}$
$P E + K E = 0$
$\frac{1}{2} {m}_{e} {v}_{e}^{2} = 2.3 \times {10}^{-} 13$
or ${v}_{e} = \sqrt{\frac{2.3 \times {10}^{-} 13 \times 2}{9.11 \times {10}^{-} 31}}$
$\approx 7.1 \times {10}^{8} m / s$

We see that it is more than the speed of light. As such electron can never escape out of nucleus and would be bound as it can never have velocity more than the speed of light.

We obtain incompatible results in both cases. Therefore, electron can not be found in the nucleus.

*Even if an electron is there, we will not be able to locate one for sure.