The uncertainty in the position of an electro | vedclass

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Question

(C) According to Heisenberg's uncertainty principle, $\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Given, $\Delta x = 100 \ pm = 100 	imes 10^{-12}

Vedclass · Accepted Answer

$0.527 	imes 10^{-24}$

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle, $\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Given, $\Delta x = 100 \ pm = 100 	imes 10^{-12} \ m = 10^{-10} \ m$.$h = 6.626 	imes 10^{-34} \ J \ s$.Substituting the values, $\Delta p \geq \frac{h}{4\pi \Delta x}$.$\Delta p \geq \frac{6.626 	imes 10^{-34}}{4 	imes 3.14159 	imes 10^{-10}}$.$\Delta p \geq \frac{6.626 	imes 10^{-34}}{12.566 	imes 10^{-10}}$.$\Delta p \geq 0.527 	imes 10^{-24} \ kg \ m \ s^{-1}$.Thus, the correct option is $C$.

The uncertainty in the position of an electron $(\Delta x)$ is approximately $100 \ pm$. Calculate the uncertainty in momentum $(\Delta p)$ of the electron in $kg \ m \ s^{-1}$. $[h = 6.626 \times 10^{-34} \ J \ s]$

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