Question

The mass of electron is $9.11\times {10}^{-31}$ kg.Planck's constant is $6.626\times {10}^{-34}$ Js then 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: (D)

According to Heisenberg's uncertainity principle,

$\Delta x.\Delta p\ge \frac{h}{4\pi }$

Here,

$h=$ Planck's constant

$\Delta x=$ Uncertainity in position

$\Delta p=$ Uncertainity in momentum

As we know that,

$\Delta p=m.\Delta v$

$\therefore \left(m\Delta v\right).\Delta x\ge \frac{h}{4\pi }$

$\Rightarrow \Delta v.\Delta x\ge \frac{h}{4\pi m}$

Given that-

$h=6.626\times {10}^{-34}J-s$

$m=9.11\times {10}^{-31}kg$

$\Delta x={10}^{-10}m$

$\Delta v=$ uncertainity in velocity $=?$

$\therefore {10}^{-10}.\Delta v\ge \frac{6.626\times {10}^{-34}}{4\times \left(\frac{22}{7}\right)\times 9.1\times {10}^{-31}}$

$\Rightarrow \Delta v\ge \frac{6.626\times {10}^{-34}\times 7}{4\times 22\times 9.1\times {10}^{-31}\times {10}^{-10}}$

$\to \Delta v\ge 5.79\times {10}^{5}m/s$

Hence the uncertainity in velocity is $5.79\times {10}^{5}m/s$.