Consider a proton moving directly towards a nucleus. The nucleus has a charge of
ID: 1335298 • Letter: C
Question
Consider a proton moving directly towards a nucleus. The nucleus has a charge of 36 qe, where qe is the charge of the proton. If the kinetic energy of the proton is 30 keV (30 x 103 electron Volts) when it is a distance di = 2.1 x 10-11 m from the nucleus, what is the kinetic energy when the proton is a distance df = di/4 from the nucleus? Assume that the nucleus does not move during this time. Give your answer in keV to three digits. Do not include units in your answer. In terms of unit conversions, note that the standard units for energy are Joules, and that 1 eV = 1.6 x 10-19 J (the change in potential energy a proton or electron when it moves through a potential diference of one Volt). So, 1 keV = 1.6 x 10-16 J.
Explanation / Answer
Proton while moving towards the nucleus will move against the electric force will loos KE and gain PE. We use conservation of energy.
The proton has a charge of qe and nucleus 36qe
Inital distance di = 2.11 x 10-11 m and final distance = di/4
The Potential at distance r from a charge Q = k* Q/r (where contant k = 9x109 N-m2 /C2 )
Thee intial and Potential difference between the final poistions of the proton is
= k * 36 qe * (4/di - 1/di) = k *108qe/di = 9 x109 x 108 x 1.6 x 10-19 /(2.11x10-11) (charge of proton = 1.6 x 10-19 C)
= 737 V
When the proton passes throguh a PD of 1 V the gain in PE is 1ev
Hence the PE gained by the proton = 737 ev
Final KE = 30000 -737 = 29263 ev
Final KE in Kev = 29.263
Note: 30 Kev is given as 30 x 103 ev, which is in correct.
If we use the above formula
Final KE of the Proton = (30 x 103 - 737 )/103 = 22.84