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A measurement of an electron's speed is `v = 2.0 x 10^6 m/s` and has an uncertainty of `10%.` What is the minimum uncertainty in its position? (`h = 6.626 x 10^-34 J*s,`  `m_el = 9.11 x 10^-31 kg`). Show steps. I got `.91 nm,` is it correct?

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Borys Shumyatskiy eNotes educator | Certified Educator

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Hello!

The uncertainty principle (one of them) states that it is impossible to exactly measure the speed and the position of a body. This inexactness is called uncertainty. Of course it is significant only for very small particles of matter, for example for electrons.

The main formula for the speed-position uncertainty is

`Delta p*Delta x gt= bar h/2,`

where `Delta p` is the uncertainty of the momentum, `Delta x` is the uncertainty of the position and `bar h` is the reduced Plank's constant  `h/(2pi).` ` `

The momentum is the mass multiplied by the speed and `10%` is `0.1.` Thus we obtain the inequality

`0.1*m_(el)*v*Delta x gt= h/(4pi),`

and therefore

`Delta x gt= h/(4pi)*10/(m_(el)*v) =(6.626*10^(-34))/(4pi)*10/(9.11*10^(-31)*2*10^6)=`

`=(6.626)/(4pi*9.11*2)*10^(-8) approx0.0289*10^(-8) = 2.89*10^(-10) (m).`

In nanometers it is `0.289 nm.`

The answer: the minimum uncertainty in position is about 0.289 nm.

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