Question

In an atom, an electron is moving with a speed of $600m/s$ with an accuracy of 0.005 %. Certainity with which the position of the electron can be located is

($h=6.6\times {10}^{-34}kg{m}^2{s}^{-1}$,mass of electron, $e^m=9.1\times {10}^{-31}kg$)

• A. $5.10\times {10}^{-3}m$
• B. $1.92\times {10}^{-3}m$
• C. $3.84\times {10}^{-3}m$
• D. $1.52\times {10}^{-4}m$

Answer 1

The correct option is B

$1.92\times {10}^{-3}m$

According to Heisenberg uncertainty principle.

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

Here $\Delta v$ = $\frac{600\times 0.005}{100}=0.03$

So, $\Delta x=\frac{6.6\times {10}^{-34}}{4\times 3.14\times 9.1\times {10}^{-31}\times 0.03}$

= $1.92\times {10}^{-3}meter$