Question

If the position of the electron is mesured within an accuracy of +0.002 nm, calculate the uncertainity in the momentum of the electton

Answer 1

Answer:

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Answer 2

Answer:

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.