How much work is done in moving Avogardo's number of electrons from an inital point where the electric potential is 9.00V to a point where the potential is -5V?
This is how I attempted the question
Delta V = V2 - V1 = -5 - 9 = -14V
W = U = Vqe
2007-11-24 06:11:00 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
e = -1.6 x10^-19) C the charge of the electron
N = 6.02 x 10^23 electrons
V = -14V
Delta V = V2 - V1 = -5 - 9 = -14V You're correct.
W = Vqe = -14V( 6.02 x 10^23)(-1.6 x10^-19) = 1,348,480 J
Please check the computation with your calculator.
2007-11-24 06:20:40 · answer #1 · answered by rene c 4 · 0⤊ 0⤋
Looks good to me.
though
W= I*U*t
Charge: Q=I*t
so
W=Q*U
now you need to know what the charge of 1 electron is and multiply that by avogadro's number to get Q
here: Q=Q(e-)*Avogadro
Q(e-)=1.602*10^(-19) C
Avogadro= 6.023*10^23 mol^-1
2007-11-24 14:24:48 · answer #2 · answered by klimbim 4 · 0⤊ 0⤋