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Question

If uncertainty in momentum of an electron is $$1\times 10^{-5}kg\ ms^{-1}$$, then what will be uncertainty in position?

Solution
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According to Heisenberg's equation:
$$\Delta x\times \Delta p=\dfrac{h}{4\pi}$$

where $$\Delta x \ \& \ \Delta p$$ are uncertainity in position and momentum of electron respectively

$$\implies$$ Uncertainity in position = $$=6.63\times 10^{-34}/4\times 3.14\times (1\times 10^{-5})$$ $$=5.28\times 10^{-30}m$$

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