# Bohr model find orbital radius for electron

#### awertag

1. Homework Statement

In the Bohr model, the electron is imagined to move in a circular orbit about a stationary proton. If the speed of the electron were 8.9e5 m/s, what would be the corresponding orbital radius?

2. Homework Equations

3. The Attempt at a Solution Fe=ma
Fe=ma
a=Fe/m
(v^2)/r=(kqq/r)/m
v^2rm=kqq
(8.9e5)(r)(9.11e-31)=(9e9)(1.6e-19)(-1.6e-19)
r=-2.842e-4m

I hope you can help!!

--aweg

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#### collinsmark

Homework Helper
Gold Member
(v^2)/r=(kqq/r)/m
'Quick note regarding Coulomb's law,

$$\vec F = k \frac{q_1 q_2}{r^2} \hat r.$$

If q1 and q2 are of the same sign, multiplying them together gives a positive sign. If the result is positive, it means that the force acting on the second particle points from the first particle toward the second particle. This is because the unit vector $$\hat r$$ is defined such that the force on a given particle points from the other particle toward the given particle. [Edit: This is opposite the way Newton's law of gravity defines the unit vector, by the way.]

In other words, ignoring the vector notation for the moment, if

$$F = k \frac{q_1 q_2}{r^2}$$

is positive, it means the charges repel. If F is negative, the charges attract (the negative sign implies that the force is in the opposite direction; i.e. the force on a given particle points toward the other particle, instead of away from it).

In your answer you came up with a negative value for r, and the above explanation is where it came from.

But you should rethink what it all means before carrying the sign all the way through to r. (Hint: your final r needs to be positive.)
v^2rm=kqq
(8.9e5)(r)(9.11e-31)=(9e9)(1.6e-19)(-1.6e-19)
r=-2.842e-4m
You forgot to square something.

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#### awertag

ok i've got it now. thanks so much!! have a good day :)

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