Question

What is the maximum precision with which the momentum of an electron can be known if the uncertainity in the position of electron is $\pm 0.001\stackrel{\circ }{A}$? Will there be any problem in describing the momentum if it has a value $\frac{h}{2\pi a_0}$, where $a_0$ is Bohr's radius of first orbit, i.e., 0.529 $\stackrel{\circ }{A}$.

• A. The uncertainity in momentum seems to be about $263.5$ times as large as the momentum itself is.
• B. The uncertainity in momentum seems to be about $243.5$ times as large as the momentum itself is.
• C. The uncertainity in momentum seems to be about $273.5$ times as large as the momentum itself is.
• D. none of the above

Answer 1

The correct option is A The uncertainity in momentum seems to be about $263.5$ times as large as the momentum itself is.

from Heisenberg uncertainity principle
$\Delta x.\Delta p\ge \frac{h}{4\pi }$
$\therefore \Delta x=0.001\stackrel{\circ }{A}={10}^{-3}\stackrel{\circ }{A}={10}^{-13}m\therefore \Delta p=\frac{6.6\times {10}^{-34}}{4\times 3.14\times {10}^{-13}}=5.27\times {10}^{-22}Ns$
Now if the given momentum = $\frac{h}{2\pi a_0}=\frac{6.6\times {10}^{-34}}{2\times 3.14\times 0.529\times {10}^{-10}}=2\times {10}^{-24}Ns$
The uncertainity in momentum seems to be about $\left(\frac{5.27\times {10}^{-22}}{2\times {10}^{-24}}\right)=263.5$ times as large as the momentum itself is. Because of this reason, the concept of Bohr's orbit has been replaced by probabilities of locating electron cloud.