If Delta x is the uncertainty in position and | vedclass

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

Question

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

Vedclass · Accepted Answer

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

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 $\Delta x$ is equal to the uncertainty in velocity $\Delta v$,so $\Delta x = \Delta v$.Since $\Delta p = m \Delta v$,we can substitute $\Delta v = \Delta x$ into the uncertainty relation:$\Delta x \cdot (m \Delta x) = \frac{h}{4 \pi}$$m (\Delta x)^2 = \frac{h}{4 \pi}$$(\Delta x)^2 = \frac{h}{4 \pi m}$$\Delta x = \frac{1}{2} \sqrt{\frac{h}{\pi m}}$Since $\Delta p = m \Delta v$ and $\Delta v = \Delta x$,then $\Delta p = m \Delta x$.Substituting the value of $\Delta x$:$\Delta p = m \cdot \frac{1}{2} \sqrt{\frac{h}{\pi m}}$$\Delta p = \frac{1}{2} \sqrt{\frac{m^2 h}{\pi m}}$$\Delta p = \frac{1}{2} \sqrt{\frac{m h}{\pi}}$

If $\Delta x$ is the uncertainty in position and $\Delta v$ is the uncertainty in velocity of a particle are equal,the correct expression for uncertainty in momentum for the same particle is

Similar Questions

The Heisenberg uncertainty principle can be applied to :

Calculate the uncertainty in the position of a particle when the uncertainty in momentum is $1 \times 10^{-3} \, g \, cm \, sec^{-1}$.

Explain the graph of $\psi(r)$ and $\psi^2(r)$ with distance $r$ from the nucleus.

The uncertainty in momentum of an electron is $1 \times 10^{-5} \ kg \ m/s$. The uncertainty in its position will be $(h = 6.63 \times 10^{-34} \ Js)$.

The Bohr model of the atom is contradicted by

Vedclass Test Series

Exam Paper Generator

Online Exam Module