Question

A microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1A^{\circ }$. What is the uncertainty involved in the measurement of its velocity?

Answer 1

Uncertainty involved in the measurement of velocity:

$\text{Given}:\Delta x=0.1\times {10}^{-10}m$

We know,according to Heisenberg uncertainty principle
$\Delta x.\Delta v=\frac{h}{4\pi m}$
$\Delta v=\frac{h}{4\pi \Delta xm}$

$\Delta v=\frac{6.626\times {10}^{-34}Js}{4\times 3.14\times (0.1\times {10}^{-10}m)\times (9.11\times {10}^{-31kg})}$

$\{\text{mass ofelectron}=9.11\times {10}^{-31}kg\}$

$\Delta v=0.579\times {10}^{7}m{s}^{-1}$

$\Delta v=5.79\times {10}^{6}m{s}^{-1}$ $\{1J=1kg{m}^2{s}^{-2}\}$

Final Answer:uncertatainty involved in the measurement of electron velocity is $5.79\times {10}^{6}m{s}^{-1}$