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grandymaker [24]
3 years ago
12

What is the uncertainty in position of an electron of an atom if there is t 2.0 x 10' msec uncertainty in its velocity? Use the

reduced Planck's constant and electron mass 9.19 x 103 kg.
Engineering
1 answer:
maw [93]3 years ago
4 0

Answer:

18931.4

Explanation:

Given : velocity of the electron = 2.0 \times10

            mass of the electron = 9.19\times 103

we know that reduced planks constant, h = 6.5821\times 10^{-16} eV s

We know from uncertainity principle,

\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}

\Delta \textup{x} = \frac{h}{\dot{m}\times \Delta \textup{v}}\Delta \textup{x} = \frac{6.5821\times 10^{-16}}{9.19\times 103\times 2.0\times 10}

\Delta \textup{x} = 18931.4 m

Hence, uncertainty in position of the electron is 18931.4

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