Asked by Anonymous
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
Answered by
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.