An electron has been accelerated by a potential difference of 100 V. If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty (Δp/p × 100) of the electron?
A) 1%
B) 2%
C) 10%
D) 10%
E) 5%

Question

An electron has been accelerated by a potential difference of 100 V. If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty (Δp/p × 100) of the electron?
A) 1%
B) 2%
C) 10%
D) >>10%
E) 5%

Quizplus · Accepted Answer

The de Broglie wavelength of the electron is given by λ = h/p, where h is Planck's constant and p is the momentum of the electron. The uncertainty in the electron's position is given by Δx = 1 nm. The uncertainty in the electron's momentum is given by Δp = h/Δx. The percent uncertainty in the electron's momentum is given by Δp/p × 100 = (h/Δx) / (h/p) × 100 = (p/Δx) × 100. The momentum of the electron is given by p = √(2meV), where me is the mass of the electron and V is the potential difference through which the electron has been accelerated. Substituting this into the expression for the percent uncertainty in the electron's momentum, we get Δp/p × 100 = (√(2meV)/Δx) × 100. Substituting the given values, we get Δp/p × 100 = (√(2 × 9.11 × 10^-31 kg × 100 V)/1 nm) × 100 = 1%.

An Electron Has Been Accelerated by a Potential Difference of 100