According to Heisenberg's uncertainty princip | vedclass

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

Question

(C) According to Heisenberg's uncertainty principle,the relation is $\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m 	imes \Delta v

Vedclass · Accepted Answer

$5.8 	imes 10^{-5} \ m^2 \ s^{-1}$

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle,the relation is $\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m 	imes \Delta v$,the equation becomes $\Delta x 	imes \Delta v 	imes m = \frac{h}{4\pi}$.Therefore,the product of uncertainties in position and velocity is $\Delta x 	imes \Delta v = \frac{h}{4 \pi 	imes m}$.Given: $h = 6.63 	imes 10^{-34} \ kg \ m^2 \ s^{-1}$,$m = 9.1 	imes 10^{-31} \ kg$,and $\pi = 3.14$.Substituting the values: $\Delta x 	imes \Delta v = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14 	imes 9.1 	imes 10^{-31}}$.$\Delta x 	imes \Delta v = \frac{6.63 	imes 10^{-34}}{114.296 	imes 10^{-31}} \approx 0.058 	imes 10^{-3} \ m^2 \ s^{-1} = 5.8 	imes 10^{-5} \ m^2 \ s^{-1}$.

According to Heisenberg's uncertainty principle,the product of uncertainties in position and velocity for an electron of mass $9.1 \times 10^{-31} \ kg$ is

Similar Questions

If the uncertainty in velocity is $\frac{1}{2m} \sqrt{\frac{h}{\pi}}$,then the ratio of uncertainty in position and momentum is (in $: 1$)

The wave mechanical model of atom is based upon

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

The uncertainty principle and the concept of wave nature of matter were proposed by ...... and ...... respectively.

Vedclass Test Series

Exam Paper Generator

Online Exam Module