Consider a proton moving directly towards a nucleus. The nucleus has a charge of

Question

Consider a proton moving directly towards a nucleus. The nucleus has a charge of 38 qe, where qe is the charge of the proton.

If the kinetic energy of the proton is 40 keV (40 x 103 electron Volts) when it is a distance di = 1.5 x 10-11 m from the nucleus, what is the kinetic energy when the proton is a distance df = di/5 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

Gain in potential energy = Loss in kinetic energy as the proton moves towards the nucleus.

PEf - PEi = KEi - KEf

Q1 = 38 * qe = 38 * 1.60217662 × 10-19 = 6.0883 * 10-18

Q2 = 1.6022 × 10-19

PEi = k * Q1 * Q2 / (di * 1.6 x 10-19) in eV

PEi = 8.9876 * 109 * 6.0883 * 10-18 * 1.6022 × 10-19 / (1.5 x 10-11 * 1.6 x 10-19)

PEi = 3.653 * 103 eV

df = di / 5 => PE will be 5 times as it is inversely proportional to distance.

PEf = 5 * PEi = 18.265 * 103 eV

Substituting in PEf - PEi = KEi - KEf

18.265 * 103 eV - 3.653 * 103 eV = 40 * 103 - KEf

14.612 * 103 eV = 40 * 103 - KEf

KEf = 40 * 103 - 14.612 * 103 eV

KEf = 25.3882  * 103 eV = 25.3882 KeV

Therefore, the kinetic energy when the proton is at a distance of df is 25.3882 KeV

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