If uncertainty in position and momentum of an | vedclass

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

Question

(A) According to Heisenberg's uncertainty principle: $\Delta x 	imes \Delta p \ge h / (4 \pi)$.Given that the uncertainty in position and momentum ar

Vedclass · Accepted Answer

$1 / (2 m) \sqrt{h / \pi}$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle: $\Delta x 	imes \Delta p \ge h / (4 \pi)$.Given that the uncertainty in position and momentum are equal: $\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = h / (4 \pi)$.Taking the square root on both sides: $\Delta p = \sqrt{h / (4 \pi)} = 1 / 2 \sqrt{h / \pi}$.Since $\Delta p = m \Delta v$,we have $m \Delta v = 1 / 2 \sqrt{h / \pi}$.Therefore,the uncertainty in velocity is $\Delta v = 1 / (2 m) \sqrt{h / \pi}$.

If uncertainty in position and momentum of an electron are equal,then uncertainty in its velocity is

Similar Questions

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 the polar coordinates of the Schrodinger equation,the angle between the radius vector and the $Z$-axis is known as ..........

Write down Heisenberg's uncertainty principle.

When the uncertainty in the position of a moving particle is $0$,then the uncertainty in momentum $(p)$ is equal to:

Assertion $(A)$: The probability of finding an electron in a small volume around a point $(x, y, z)$ at a distance $r$ from the nucleus is proportional to $\psi^2$.
Reason $(R)$: Subatomic particles possess both particle and wave nature.

Vedclass Test Series

Exam Paper Generator

Online Exam Module