Uncertainty in position and momentum are equa | vedclass

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Question

(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given that the uncertainty in position and momentum

Vedclass · Accepted Answer

$\frac{1}{2m}\sqrt{\frac{h}{\pi}}$

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given that the uncertainty in position and momentum are equal,$\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = \frac{h}{4\pi}$.Therefore,$\Delta p = \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}$.Since $\Delta p = m \cdot \Delta v$,the uncertainty in velocity is $\Delta v = \frac{\Delta p}{m}$.Substituting the value of $\Delta p$: $\Delta v = \frac{1}{m} \cdot \frac{1}{2} \sqrt{\frac{h}{\pi}} = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.

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

Similar Questions

The Heisenberg uncertainty principle can be applied to :

In an atom,an electron is moving with a speed of $600 \, m/s$ with an accuracy of $0.005 \%$. The certainty with which the position of the electron can be located is $(h = 6.6 \times 10^{-34} \, kg \, m^2 s^{-1}, m_e = 9.1 \times 10^{-31} \, kg)$:

If the uncertainty in position is of the order of $1 \, \mathring{A}$,what will be the uncertainty in velocity of a cricket ball weighing $150 \, g$?

Uncertainty in the position of an electron (mass $= 9.1 \times 10^{-31} \ kg$) moving with a velocity $300 \ ms^{-1}$ accurate up to $0.001 \ \%$ will be $(h = 6.63 \times 10^{-34} \ J \cdot s)$.

If the uncertainty in momentum of a particle is twice the uncertainty in its position,what will be the uncertainty in the position of the particle?

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