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(A) According to Heisenberg's Uncertainty Principle,the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is gi

Vedclass · Accepted Answer

$32 	imes 10^5$

Vedclass · Answer

(A) According to Heisenberg's Uncertainty Principle,the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is given by: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$ Since the uncertainty in position $(\Delta x)$ is the same for both the electron and the helium atom,and the value of $h/4\pi$ is a constant,the uncertainty in momentum $(\Delta p)$ must also be the same for both particles. Given that the uncertainty in the momentum of the electron is $\Delta p_e = 32 	imes 10^5$,the uncertainty in the momentum of the helium atom must also be $\Delta p_{He} = 32 	imes 10^5$. Therefore,option $A$ is correct.

The uncertainty in the position of an electron and a helium atom is the same. If the uncertainty in the momentum of the electron is $32 \times 10^5$,then the uncertainty in the momentum of the helium atom is:

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