The mass of an electron is 9.11 times 10^{-31 | vedclass

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

Question

(B) According to Heisenberg's uncertainty principle:$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$Since $\Delta p = m \cdot \Delta v$,the formula beco

Vedclass · Accepted Answer

$5.79 	imes 10^6 \ m \ s^{-1}$

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle:$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$Since $\Delta p = m \cdot \Delta v$,the formula becomes:$\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4 \pi}$$\Delta v \geq \frac{h}{4 \pi \cdot m \cdot \Delta x}$Given:$\Delta x = 0.1 \ \mathring{A} = 0.1 	imes 10^{-10} \ m = 10^{-11} \ m$$m = 9.11 	imes 10^{-31} \ kg$$h = 6.626 	imes 10^{-34} \ J \ s$$\pi = 3.14$Substituting the values:$\Delta v = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14 	imes 9.11 	imes 10^{-31} 	imes 10^{-11}}$$\Delta v = \frac{6.626 	imes 10^{-34}}{114.46 	imes 10^{-42}}$$\Delta v \approx 0.05789 	imes 10^8 \ m \ s^{-1} = 5.79 	imes 10^6 \ m \ s^{-1}$

The mass of an electron is $9.11 \times 10^{-31} \ kg$,and the Planck constant is $6.626 \times 10^{-34} \ J \ s$. The uncertainty involved in the measurement of velocity within a distance of $0.1 \ \mathring{A}$ is:

Similar Questions

Which two developments remove the drawbacks of Bohr's model?

"The position and velocity of a small particle like an electron cannot be simultaneously determined." This statement is:

The wave mechanical model of atom is based upon

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

The Bohr model of an atom is contradicted by

Vedclass Test Series

Exam Paper Generator

Online Exam Module