Asked by Todd
An alpha particle (which has two protons) is sent directly toward a target nucleus containing 80 protons. The alpha particle has an initial kinetic energy of 0.79 pJ. What is the least center-to-center distance the alpha particle will be from the target nucleus, assuming the nucleus does not move? The charge of a proton is 1.602 x 10-19 C.
Answers
Answered by
drwls
The potential energy due to the Coulomb replsion force is
-k Q1 Q2/R
at separation distance.
k is the Coulomb constant and Q1 and Q2 are the two positive nuclear charges. When the nuclei are at closest separation distance d, the relative velocity is zero. You can ignore the effect of the 80 electrons around the larger nucleus.
Initial kinetic energy = potential energy change
0.79*10^-12 J = k Q1 Q2/d = 2*80 k e^2/d
Solve for d
-k Q1 Q2/R
at separation distance.
k is the Coulomb constant and Q1 and Q2 are the two positive nuclear charges. When the nuclei are at closest separation distance d, the relative velocity is zero. You can ignore the effect of the 80 electrons around the larger nucleus.
Initial kinetic energy = potential energy change
0.79*10^-12 J = k Q1 Q2/d = 2*80 k e^2/d
Solve for d
Answered by
Caleb
He has the right equation, but the wrong exponents.
Equation = (r = (KQ)/K0)
K = 9 * 10^9
Q = (2*e*#protons*e)
K0 = initial ke * 10^-12
Example:
(K*(2*1.602E-19*86*1.602E-19))/.39E-12
= 1.017E-13m
Equation = (r = (KQ)/K0)
K = 9 * 10^9
Q = (2*e*#protons*e)
K0 = initial ke * 10^-12
Example:
(K*(2*1.602E-19*86*1.602E-19))/.39E-12
= 1.017E-13m
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