Asked by Jake
) What is the minimum uncertainty in the velocity of an electron that is known to be somewhere between 0.050 nm and 0.10 nm from a proton?
Answers
Answered by
Damon
delta x = .05 *10^-9 m
delta x * delta p = (1/2) hbar = .5*10^-34
delta p = (.5*10^-34)/.05*10^-9)
= 10 * 10^-25 kg m/s
mass of electron * delta v = 10*-24
delta v = (10*10^-25)/(9*10^-31)
delta v = 1.1 * 10^6 m/s
delta x * delta p = (1/2) hbar = .5*10^-34
delta p = (.5*10^-34)/.05*10^-9)
= 10 * 10^-25 kg m/s
mass of electron * delta v = 10*-24
delta v = (10*10^-25)/(9*10^-31)
delta v = 1.1 * 10^6 m/s
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.