Question

An electron has a speed of 4$\times {10}^5$ m/s accurate to point 0.01% with what fundamental accuracy can we locate the position of the electron

Answer 1

We have,

$Speed4\times {10}^5$

Acoording to the Uncertainity principle,

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

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

$\Delta v=4\times {10}^{5}m/s$ (accurate upto $0.01\%$)

$=\frac{(100-0.01)}{100\times 4\times {10}^{5}m/s}$

$\frac{99.99}{4\times {10}^{7}m/s}$

$=24.99\times {10}^{-7}m/s$

$m=9.1\times {10}^{-31}kg,h=6.626\times {10}^{-34}kg{m}^2/s,\pi =3.14$

Therefore,

$\Delta x=\frac{6.626\times {10}^{-34}kg{m}^2/s}{4\times 3.14\times 24.99\times {10}^{-7}m/s\times 9.1\times {10}^{-31}kg}$

$\frac{6.626\times {10}^{-34}kg{m}^2/s}{2857.11\times {10}^{-39}}$

$=0.0023\times {10}^5$