A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1A°. What is the uncertaintiy involved in the measurement of its velocity?

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$0.79 \times 10^{6} {ms}^{- 1}$

$5.79 \times 10^{16} {ms}^{- 1}$

$5.79 \times 10^{6} {ms}^{- 1}$

$5.79 {ms}^{- 1}$

answer is C.

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Detailed Solution

$∆ x \times m ∆ v = \frac{h}{4 π}$

$∆ v = \frac{h}{4 π \times ∆ x \times m}$

m =9.11 $\times$ 10^-31kg ; $∆ x$ =0.1A⁰ :h =6.625 $\times$ 10^-34kg m² s^-1

$∆ v = \frac{6.625 \times 10^{- 34}}{4 \times 3.142 \times (10^{- 10}) \times 9.11 \times 10^{- 31}}$

$∆ v = 5.79 \times 10^{6} m s^{- 1}$

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