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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 .....

Similar Questions

Uncertainty in position and momentum are equal. Uncertainty in velocity is :-

The uncertainty in the location of a circulating electron is equal to its de Broglie wavelength. The minimum percent error in its measurement of velocity under these circumstances will be approximately:

If the position of the electron was measured with an accuracy of $\pm 0.002 \ nm$,the uncertainty in the momentum of it would be (in $kg \ ms^{-1}$) $(h=6.626 \times 10^{-34} \ J \ s)$.

Which one of the following conditions is incorrect for a well-behaved wave function $(\psi)$?

The uncertainty in the position and velocity of a particle are $10^{-10} \ m$ and $5.27 \times 10^{-24} \ ms^{-1}$ respectively. The mass of the particle is ............. $kg$. $(h = 6.625 \times 10^{-34} \ JS)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module