Reunion of proton and electron!

Classical Mechanics Level 3

If the potential energy between a proton and an electron is given by $\text|{U}|=\dfrac{te^2}{2R^3}$ , where $e$ is the charge of the electron and $R$ is the radius of the Boh'r Orbit.

Then which of the following is the correct expression for $R$ ?

Details and Assumptions:

$\rightarrow$ $t$ is a constant.

$\rightarrow$ $h$ is Planck's constant.

$\rightarrow$ $n$ is principal quantum number.

\dfrac{2{\pi}^2te^2m}{n^2h^2}

\dfrac{6{\pi}^2te^2m}{n^2h^2}

\dfrac{te^2m}{h^2}

\dfrac{4{\pi}^2te^2m}{n^2h^2}

1 solution

Skanda Prasad
Oct 17, 2017

Note: In all the equations, I have considered only magnitude part without giving importance to signs.

Hey, your answer is wrong. The correct answer is the first option.

Rahul Kumar - 3 years, 7 months ago

Why bro? Please send your solution. I won't understand unless you explain. And please point out the mistake in my solution. In which step have I gone wrong?

Skanda Prasad - 3 years, 7 months ago

You have to derive net electrostatic force by differentiating potential wrt radial distance. In this way u will get a force field which is proportional to r^-4 because this is a hypothetical situation as the question suggests. You have already assumed that force is proportional to r^-2 which would give potential proportional to r^-1 but in the question it is proportional to r^-3

Rahul Kumar - 3 years, 7 months ago

@Rahul Kumar – Ah, yes bro...6pi is correct

Skanda Prasad - 3 years, 7 months ago

@Skanda Prasad – Yeah bro.....correct the answer of the question please

Rahul Kumar - 3 years, 7 months ago

Reunion of proton and electron!

1 solution

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