In a He atom,the uncertainty in the position | vedclass

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Question

(B) According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Given that the uncertainty in position $\Delta x$ i

Vedclass · Accepted Answer

$32 	imes 10^5$

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Given that the uncertainty in position $\Delta x$ is equal to the uncertainty in momentum $\Delta p$ for the electron,i.e.,$\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = \frac{h}{4\pi}$,so $\Delta p = \sqrt{\frac{h}{4\pi}}$.Since the question states the uncertainty in position of the $He$ atom is equal to the uncertainty in its momentum,and the numerical value provided for the electron's uncertainty is $32 	imes 10^5$,the value for the $He$ atom remains the same under the given condition.Therefore,$\Delta p_{He} = 32 	imes 10^5$.

In a $He$ atom,the uncertainty in the position of an $e^-$ is equal to the uncertainty in its momentum $\Delta p$. If the value of $\Delta p$ for the $e^-$ is $32 \times 10^5$,then the value of $\Delta p$ for the $He$ atom is .....

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