1. In Dom\'s universe (not our universe), the magnitude of the force of attracti
ID: 1777998 • Letter: 1
Question
1. In Dom's universe (not our universe), the magnitude of the force of attraction between the proton and the electron in a hydrogen atom is: F-k2 where r is the radius of the circular orbit, and k is a constant that contains information about the magnitude of the electric charge, e, of both particles. .Assume the proton is fixed in place and the electron is in a circular orbit of radius r1. The electron "jumps" to a new, smaller radius r2. What is the change in the total energy of the system? Hint: Use dynamics to calculate the kinetic energy for the electron in the initial and final orbits, and then calculate the potential energy in both orbits (you will need to derive an expression for the potential energy). Express your answer in terms of the initial and final orbital radii r and r2, k, and numerical constants. r2 electro proton r1Explanation / Answer
F = m_e v^2 / r
k r^2 = m_e v^2 / r
KE = m_e v^2 /2 = k r^3 / 2
U = integral of -Fdr = - k r^3 / 3
total energy = U + KE = 0.167 k r^3
change in energy = 0.167 k (r2^3 - r1^3) ....Ans
KEi = k r1^3 / 2
KEf = k r2^3 / 2
PEi = - k r1^3 / 3
PEf = - k r2^3 / 3
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