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. $5.79\times 10^{6}m{s}^{-1}$
• B. $5.79\times 10^{7}m{s}^{-1}$
• C. $5.79\times 10^{8}m{s}^{-1}$
• D. $5.79\times 10^{5}m{s}^{-1}$

Answer 1

Answer: (A)

By Heisenberg's uncertainty principle
$\Delta x\times \Delta px\ge \frac{h}{4\pi }$
or $\Delta x\times \Delta \left(m{v}_x\right)\ge \frac{h}{4\pi }$
$\Delta x\times \Delta v_x\ge \frac{h}{4\pi m}$
$\Delta p=$ 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,
$\Delta v\times 0.1\times 10^{-10}=\frac{6.626\times 10^{-34}}{4\times 3.14\times 9.11\times 10^{-31}}$
$\Delta v=\frac{6.626\times 10^{-34}}{4\times 3.14\times 9.11\times 10^{-31}\times 0.1\times 10^{-10}}m{s}^{-1}$
$=5.785\times 10^{6}m{s}^{-1}$
$=5.79\times 10^{6}m{s}^{-1}$