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if the position of the electron is measured within...

Question:
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/4π_m × 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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Question:
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/4π_m × 0.05 nm, is there any problem in defining this value.

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Question: 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.

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Question:
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/4π_m × 0.05 nm, is there any problem in defining this value.