Q

# If the position of the electron is measured within an accuracy of +- 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h / 4 pi m x 0.05 nm, is there any problem in defining this

2.61    If the position of the electron is measured within an accuracy of $\pm 0.002\ \textup{nm}$, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $h/4\pi_m\times0.05 \ \textup{nm}$, is there any problem in defining this value.

Views

We have given the uncertainty in position, i.e., $\triangle x= \pm 0.002\ nm = 2\times10^{-12} m$.

According to Heisenberg's Uncertainty Principle:

$\triangle x\times \triangle p = \frac{h}{4\pi}$

Where,

$\triangle x$ is uncertainty in the position of the electron.

$\triangle p$ is uncertainty in the momentum of the electron.

Then, $\triangle p = \frac{h}{4\pi\times\triangle x}$

$\triangle p =\frac{6.626\times10^{-34}Js}{4\pi\times(2\times10^{-12}m)} =2.636\times10^{-23}Jsm^{-1}$

Or $2.636\times10^{-23}kgms^{-1}$          $\left ( 1J- 1kgms^2s^{-1} \right )$

The actual momentum of the electron:

$\frac{h}{4\pi_{m}\times0.05\ nm}= \frac{6.626\times10^{-34}Js}{4\pi\times0.05\times10^{-9} m }$

$\Rightarrow p = 1.055\times10^{-24}kg\ m/s$

Therefore, it cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.

Exams
Articles
Questions