A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the minimum uncertainty possible in its speed? h = 6.626 * 10^(-34) J * s, m_(el) = 9.11 * 10^-31 kg, 1 eV = 1.60 * 10^(-19) J). I got both .15 eV and 1.5 eV and I dunno if correct. Choices are A) 1.5 eV B) .015 eV C) 150 eV D) 15 eV E) .15 eV Wordz from old buddy previous test said "The minimum kinetic energy of the electron is closest to" not this minimum uncertainty stuff. old answer was .15 eV.

No actually 1.5 eV is correct I asked my professor but how do you get that then?

Hello!
Again, we have a deal with the uncertainty principle for a position and a momentum. Mathematically it is expressed as
Delta p*Delta x gt= h/(4 pi),
where Delta p = m*Delta v is an uncertainty in a momentum and Delta x is the uncertainty in a position.
Also we know that the kinetic energy of an electron is equal to (m_(el) v^2)/2, and it is given that the speed is the same as the minimum uncertainty of a speed. So the kinetic energy is equal to
E_k = (m_(el) v^2)/2 =(m_(el) (Delta v)^2)/2 = m_(el)/2 * (h/(4 pi) * 1/(m_(el)*Delta x))^2 =1/2 (h/(4 pi))^2 1/(m_(el)*(Delta x)^2).
All values are given, and we can compute the result in Joules (note that p means pico- means 10^(-12) ):
E_k = 1/2*(1.055*10^(-34))^2/(9.11*10^(-31)*25*10^(-20)) = (1.055)^2/(2*9.11*25)*10^(-17) approx 0.00244*10^(-17) (J).
To convert this to eV we have to divide by eV value:
E_k =0.00244*10^(-17) / (1.60*10^(-19)) = 0.00244*10^2/1.60 approx0.153 (eV).
So your hypothesis is correct, the answer is about 0.15 eV, E.