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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

If the uncertainty in position and momentum are equal,then the uncertainty in velocity is .......

Similar Questions

The measurement of the electron position is associated with an uncertainty in momentum,which is equal to $1 \times 10^{-18} \ g \ cm \ s^{-1}$. The uncertainty in electron velocity is (mass of an electron is $9 \times 10^{-28} \ g$):

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 $kg \ m \ s^{-1}$. $[h = 6.626 \times 10^{-34} \ J \ s]$

In an atom,an electron is moving with a speed of $600 \ m/sec$ with an uncertainty of $0.005\%$. The uncertainty with which the position of the electron can be located is:

If uncertainty in position and momentum are equal,then uncertainty in velocity is

Assertion : It is impossible to determine the exact position and exact momentum of an electron simultaneously.
Reason : The path of an electron in an atom is clearly defined.

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