Question

Given, the mass of electron is $9.11 \times 10^{-31}\, kg$, Planck's constant is $6.626 \times 10^{-34}\, Js$, the uncertainty involved in the measurement of velocity within a distance of $0.1\,\mathring{A} $ is

• A. $5.79 \times 10^6 \,ms^{-1}$
• B. $5.79 \times 10^7\, ms^{-1}$
• C. $5.79 \times 10^8\, ms^{-1}$
• D. $5.79 \times 10^5\, ms^{-1}$

Answer 1

Answer: (A)

By Heisenberg's uncertainty principle
$\Delta x \times \Delta p x \geq \frac{h}{4 \pi}$
or $\Delta x \times \Delta\left(m v_{x}\right) \geq \frac{h}{4 \pi}$
$\Delta x \times \Delta v_{x} \geq \frac{h}{4 \pi m}$
$\Delta p=$ uncertainty in momentum
$\Delta x=$ uncertainty in position
$\Delta v=$ uncertainty in velocity
$m=$ mass of particle
Given that,
$\Delta x=0.1\,\mathring{A} =0.1 \times 10^{-10}\, m$
$m=9.11 \times 10^{-31}\,kg$
$h=$ Planck's constant $=6.626 \times 10^{-34} \,Js$
$\pi=3.14$
Thus,
$\Delta v \times 0.1 \times 10^{-10}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31}} $
$\Delta v =\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 0.1 \times 10^{-10}} \,ms ^{-1} $
$=5.785 \times 10^{6} \,ms ^{-1}$
$=5.79 \times 10^{6} \,ms ^{-1}$