Question

(a) Assume an electron in the ground state of the hydrogen atom moves at an average speed of 5.00 × 10^6 m/s. If the speed is known to an uncertainty of 1 percent, what is the minimum uncertainty in its position? The radius of the hydrogen atom in the ground state is 5.29 × 10^−11 m. The mass of an electron is 9.1094 × 10^−31 kg.

__________× 10______ m

(b) A 0.13−kg baseball thrown at 100 mph has a momentum of 5.9 kg · m/s. If the uncertainty in measuring the mass is 1.0 × 10^−7 of the mass, calculate the uncertainty in the baseball's position.

__________× 10______ m

Answer 1

a) uncertainty principle

DX*DV = h/4*pi*m

DX = uncertainty in position = ?

DV = uncertainty in velocity = 5.00*10^6*1/100 = 5.00*10^4 m/s

h = planks constant = 6.625*10^-34 j.s

pi = 3.14

m = mass of electron = 9.1094*10^-31 kg

x*(5.00*10^4) = (6.625*10^-34)/(4*3.14*9.1094*10^-31)

DX = uncertainty in position = x = 1.158*10^-9 m

b)

DX*Dp = h/4*pi

DX = uncertainty in position = ?

Dp = uncertainty in momentum = (5.9/0.13)*1.0*10^-7 = 4.54*10^-6

h = planks constant = 6.625*10^-34 j.s

pi = 3.14

x*4.54*10^-6 = ((6.625*10^-34)/(4*3.14))

DX = uncertainty in position = x = 1.162*10^-29 m