Question 2.61 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.
Class XI Structure of Atom Page 68
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From Heisenberg’s uncertainty principle,
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
..........check the file..........
= 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
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
..........check the file..........
= 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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