Question

The uncertainty in velocity of an electron present in the nucleus of diameter ${10}^{-15}m$ hypothetically should be approximately :

• A. ${10}^{-11}m/s$
• B. ${10}^{8}m/s$
• C. ${10}^{11}m/s$
• D. $10\mathring{A}/s$

Answer 1

The correct option is C ${10}^{11}m/s$

Heisenberg's uncertainty formula = $\Delta x.m\Delta v=\frac{h}{4\pi }$

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

$\therefore \Delta v=\frac{6.634\times {10}^{-34}Js}{4\pi \left({10}^{-15}m\right).9.1\times {10}^{-31}kg}$

=$0.58\times {10}^{11}$ m/s