The figure provided shows the potential energy of a proton, q = +e, and a lead nucleus, q = +82e. If a proton is fired toward a lead nucleus from a large distance away, how much initial kinetic energy does the proton need to reach a turning point 5.00 fm from the nucleus?
Question
The figure provided shows the potential energy of a proton, q = +e, and a lead nucleus, q = +82e. If a proton is fired toward a lead nucleus from a large distance away, how much initial kinetic energy does the proton need to reach a turning point 5.00 fm from the nucleus?
Solution
The question doesn't provide a figure or any numerical values for the potential energy, so I'll have to make some assumptions to provide a general answer.
The potential energy (PE) between two charges is given by Coulomb's law:
PE = kq1q2/r
where k is Coulomb's constant (8.9910^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between them.
In this case, q1 is the charge of the proton (+e) and q2 is the charge of the lead nucleus (+82e). The distance r is given as 5.00 fm (femtometers), which is 5.00*10^-15 m.
So, the potential energy at the turning point is:
PE = kq1q2/r = (8.9910^9 Nm^2/C^2) * (+e) * (+82e) / (5.00*10^-15 m)
The proton needs at least this much kinetic energy to reach the turning point.
Note: The charge of a proton (e) is approximately 1.60*10^-19 C.
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