Question

A microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1\mathring{A}$. What is the uncertainty involved in the measurement of its velocity? ( h = 6.626 $\times {10}^{-34}$ Js, mass of electron = 9.1 $\times {10}^{-31}$ kg)

Answer 1

According to Heisenberg’s Uncertainty principle: del X del P = h/2pi

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

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

So..,

$\Delta V\ge \frac{h}{4\pi m\Delta x}$

$\Delta V\ge \frac{6.626\times 10^{-34}}{4\times 3.14\times 9.11\times 10^{-31}\times 0.1\times 10^{-10}}$

$\Delta V\ge 0.579\times 10^{7}m{s}^{-1}$

$\Delta V\ge 5.79\times 10^{6}m{s}^{-1}$