An accelerated electron has a speed of \(5 \times 10^{6}\) ms-1 with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is \(x \times 10^{-9}\) m. The value of x is ____________. (Nearest integer)
[Use mass of electron = \(9.1 \times 10^{-31}\) kg, h = \(6.63 \times 10^{-34}\) Js, \(\pi = 3.14\)]
[Use mass of electron = \(9.1 \times 10^{-31}\) kg, h = \(6.63 \times 10^{-34}\) Js, \(\pi = 3.14\)]
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Correct Answer: 58
Solution and Explanation
Heisenberg: \(\Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{3}}\).
Denominator: \(12.56 \times 9.1 \times 10^{-28} = 1.143 \times 10^{-26}\).
\(\Delta x = 5.80 \times 10^{-8}\) m \(= 58 \times 10^{-9}\) m.
So x = 58.
Answer: 58
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