Question

with the help of uncertainity principle show that electrons are not present in the atomic nucleus

Answer 1

Answer on Question #73477 - Physics / Quantum Mechanics

with the help of uncertainty principle show that electrons are not present in the atomic nucleus.

Solution:

The uncertainty principle states

$$\Delta p \Delta x \geq \frac {\hbar}{2}$$

Because $\Delta p = m\Delta v$ , we get $\Delta v \geq \frac{\hbar}{2m\Delta x}$

The numerical values of the constant

$$\hbar = 1.054 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}$$ $$m = 9.1 \times 10^{-31} \mathrm{kg}$$ $$\Delta x \sim 1 \mathrm{fm} = 10^{-15} \mathrm{m}$$

Thus

$$\Delta v \geq \frac {1.054 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 10^{-15}} = 5.5 \times 10^{10} \frac {\mathrm{m}}{\mathrm{s}}$$

So $\Delta v > c$ , where $c = 3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}$ light speed

Therefore electrons are not present in the atomic nucleus.

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