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Heisenberg's Uncertainty Principle

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Question

How can you determine the size of hydrogen atom using uncertainty principle potential energy of an electron $V = \frac{- e}{4 Π E_{0} a}$ , were 'a' is the distance of electron from the nucleus.

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Solution

$Δ x \times Δ V = \frac{h}{4 Π m} V_{n} = \frac{2.8 \times 10^{6}}{n} Δ x \times \frac{2.8 \times 10^{6}}{n} = \frac{6.626 \times 10^{- 34}}{4 Π x} Δ x \times V = \frac{h}{4 Π m} Δ x = \frac{h}{2 Π \sqrt{\frac{m^{2} V^{2}}{2}} \times 2} Δ x = \frac{h}{2 Π \sqrt{m V}} \to (1) K e = e V = e . \frac{e^{2}}{4 Π E 9} \to (11) Δ x = \frac{h}{2 Π \sqrt{m \frac{e^{2}}{4 Π E 9}}} Δ x = \frac{h}{\sqrt{4 m^{2} \frac{{m e}^{2}}{4 Π E 9}}}$

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