Based on Heisenberg's uncertainty principle,t | vedclass

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

Question

(C) According to Heisenberg's uncertainty principle:$m \Delta V \cdot \Delta x \geq \frac{h}{4 \pi}$Given:$\Delta x = 10^{-15} \ m$$m = 9.1 	imes 10^

Vedclass · Accepted Answer

$58$

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle:$m \Delta V \cdot \Delta x \geq \frac{h}{4 \pi}$Given:$\Delta x = 10^{-15} \ m$$m = 9.1 	imes 10^{-31} \ kg$$h = 6.626 	imes 10^{-34} \ Js$$\pi = 3.14$Substituting the values:$\Delta V = \frac{h}{4 \pi m \Delta x}$$\Delta V = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14 	imes 9.1 	imes 10^{-31} 	imes 10^{-15}}$$\Delta V = \frac{6.626 	imes 10^{-34}}{114.296 	imes 10^{-46}}$$\Delta V = 0.05797 	imes 10^{12} \ ms^{-1}$$\Delta V = 57.97 	imes 10^9 \ ms^{-1}$Rounding to the nearest integer,we get $58 	imes 10^9 \ ms^{-1}$.

Similar Questions

The Bohr model of an atom is contradicted by

$A$ ball weighing $10 \, g$ is moving with a velocity of $90 \, ms^{-1}$. If the uncertainty in its velocity is $5 \, \%$,then the uncertainty in its position is ....... $\times 10^{-33} \, m$. (Rounded off to the nearest integer) [Given : $h = 6.63 \times 10^{-34} \, Js$]

What is the uncertainty product $\Delta v \cdot \Delta x$ for a particle of mass $1 \ mg$? What does this imply?

The uncertainty in momentum of an electron is $1 \times 10^{-5} \ kg \ m/s$. The uncertainty in its position will be $(h = 6.62 \times 10^{-34} \ kg \ m^2/s)$

Uncertainty in the position of an electron (mass $= 9.1 \times 10^{-31} \, kg$) moving with a velocity $300 \, ms^{-1}$,with uncertainty $0.001\%$ will be :- $(h = 6.63 \times 10^{-34} \, Js)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module