Question
What is the minimum uncertainty in the position of an electron, in Angstroms, if its uncertainty in velocity is known to be 121.8 km/s
Answers
DrBob222
delta x*delta p = h/4*pi where delta p = mv
v = 121.8 km/s. Convert to m/s so
delta p = mdelta v = 9.11E-31g*121.8 km/s x (1000 m/km) = ? and substitute into
delta x = 6.626E-31/4*3.1416*delta p
Post your work if you get stuck.
v = 121.8 km/s. Convert to m/s so
delta p = mdelta v = 9.11E-31g*121.8 km/s x (1000 m/km) = ? and substitute into
delta x = 6.626E-31/4*3.1416*delta p
Post your work if you get stuck.