Question 61

If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.

Answer

From Heisenberg’s uncertainty principle,

Where,

Δx = uncertainty in position of the electron

Δp = uncertainty in momentum of the electron

Δx = 0.002nm = 2x10-12m(given)

 Therefore, Substituting the values in the expression of Δp:

Δp = h/4π Δx or

= 2.637 × 10–23 Jsm–1

Δp = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s–1)

Actual momentum = h/4πx 0.05nm

= 6.626x10-34/4 x3.14 x 5 x10-11

= 1.055 x 10-24 kg m/sec

 

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