Question

What will be the uncertainty in velocity of an electron when the uncertainty in its position is 1000 $\mathring{A}$?

• A. $5.79\times {10}^{2}m{s}^{-1}$
• B. $5.79\times {10}^{8}m{s}^{-1}$
• C. $5.79\times {10}^{4}m{s}^{-1}$
• D. $5.79\times {10}^{-10}m{s}^{-1}$

Answer 1

The correct option is A $5.79\times {10}^{2}m{s}^{-1}$

According to Heisenberg's uncertainity principle:

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

$\Delta x.\Delta \left(mv\right)=\frac{h}{4\pi }$

$\Delta v=\frac{h}{4\pi m\Delta x}$

$h=6.626\times {10}^{-34}Js$=Planck's constant

Given $m=$mass of electron$=9.1\times {10}^{-31}$ kg and $\Delta x=1000\mathring{A}={10}^{-7}$m

upon substitution we get:

$\Delta v=\frac{6.626\times {10}^{-34}}{4\times 3.14\times 9.1\times {10}^{-31}\times {10}^{-7}}$

$\Delta v=0.0579\times {10}^4=5.79\times {10}^2$m/s

So, the correct option is $A$