Question

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/4πm x 0.05 nm, is there any problem in defining this value.

Answer 1

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Answer 2

Explanation:

According to Heisenberg's uncertainity principle, relation between uncertainity in position and momentum of electron will be as follows.

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

where,  $\Delta x$ = uncertainity in position of electron

            $\Delta p$ = uncertainity in momentum of electron

                        h = plank's constant = $6.624 \times 10^{-34}$ Js

Hence, put the given values into the above formula as follows.

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

                 $\Delta p$ = $\frac{1}{0.002 nm} \times \frac{6.624 \times 10^{-34}Js}{4 \times 3.14}$

                                            = $2.637 \times 10^{-23} Jsm^{-1}$

or,                $\Delta p$ = $2.637 \times 10^{-23} Kgms^{-1}$   (as 1 J = $1 kgms^{2}. s^{-1}$)

Now, actual momentum = $\frac{h}{4 \pi \times 0.05nm}$

                                        = $\frac{6.624 \times 10^{-34}Js}{4 \times 3.14 \times 0.05nm}$

                                        = $1.055 \times 10^{-24} Kgms^{-1}$

As it is shown that the actual momentum is smaller than the uncertainity. Hence, the value cannot be defined.