Question

If the uncertainty in velocity of electron is 0.06 m/s, what is the uncertainty in position of electron? 9.7 x 10-4 m 5.8 x 10-6 m 1.44 x 10-10 m 2 x 10-3 m​

Answer 1

Given info : If the uncertainty in velocity of electron is 0.06 m/s.

To find : The uncertainty in position of electron is ...

solution : Heisenberg uncertainty principle,

∆x × ∆p = h/4π

∆p = uncertainty in linear momentum

= m∆v

given, uncertainty in velocity of electron, ∆v = 0.06 m/s

and mass of electron, m = 9.1 × 10¯³¹ kg

so, ∆p = 9.1 × 10¯³¹ × 0.06 kg m/s = 0.546 × 10¯³¹ Kgm/s

now, ∆x = h/4π∆p

= (6.63 × 10¯³⁴ Js)/(4 × 3.14 × 0.546 × 10¯³¹ kg m/s)

= (6.63 × 10¯³⁴ Js)/(6.85776 × 10¯³¹ kg m/s)

= 0.966787989 × 10¯³ m ≈ 9.7 × 10¯⁴ m

= 9.7 × 10¯⁴ m

Therefore the uncertainty in position of electron is 9.7 × 10¯⁴ m.

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