Posted by Susan on Sunday, January 13, 2008 at 10:26pm.
The hydrogen atom contains a proton, mass 1.67 1027 kg, and an electron, mass 9.11 1031 kg. The average distance between them is 5.3 1011 m. The charge of the proton is the same size, opposite sign of the electron.
a)What is the magnitude of the average electrostatic attraction between them?
F=kqq/d^2
You would have to convert from kg to C, correct? Then, plug everything into the formula.
b)What is the magnitude of the average gravitational attraction between them?
I am drawing a complete blank on the formula for this one and I cannot find it in my notes.

Physics  Count Iblis, Monday, January 14, 2008 at 9:47am
There are some mistakes in the problem. The average distance between the electron and the proton is not equal to 5.3 10^(11) m. The distance
a = 5.3 10^(11) m is the Bohr radius which and at this distance is the most likely distance for the electron to be.
The average distance is 3/2 a. In this problem you need to know the average of 1/d^2, which is given by 2/a^2. So, you need to find the charge of the electron, e, anbd evaluate:
2 k e^2/a^2
The average gravitational attraction is given by:
2 G m^2/a^2
Of course, the other mistake in this problem is that it is suggested that the average attractive force would be given by k e^2/a^2 :)
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