Question

In an atom, an electron is moving with a speed of 800 ms^-1 with an accuracy of 0.005%. Find the certainty with which the position of the electron can be located.

(h = 6.626 x 10^-34 Js, mass of electron = 9.1 x 10^-31 kg).

Answer 1

\(\Delta v\) = 800 x \(\frac{0.005}{100}\) = 0.04 ms^-1

According to Heisenberg uncertainty principle,

\(\Delta \mathrm{x}\times m\Delta v\) = \(\frac{h}{4\pi}\)

\(\therefore\) \(\Delta \mathrm{x}\) = \(\frac{h}{4\pi m.\Delta v}\)

= \(\frac{6.626\times10^{-34}kgm^2s^{-1}}{4\times3.14\times9.1\times10^{-31}kg\times0.04ms^{-1}}\)

= \(\frac{6.626\times10^{-34}}{4.572\times10^{-31}}\)

= 1.449 × 10³ m

= 1.45 × 10⁻³ m