Distance of electron from nucleus

[dinospamoni]

Homework Statement

a) Consider a hydrogenic atom consisting of a carbon-12
nucleus and a single electron. What is the wavelength of the
photon emitted when this atom drops from n = 5 to the n = 2
state?

b) According to the Bohr theory, what is the distance
between the electron and the nucleus in the n = 2 state of the
hydrogenic carbon-12 atom?

Homework Equations

$r_n = n^2 \times a_0$

where $a_0$ is the bohr radius

The Attempt at a Solution

I found the answer to part A no problem, but b is giving me trouble.

I know that the equation for the radius for excited states is for a hydrogen atom. WIth a carbon nucleus I expect it to be much smaller that the bohr radius, but I can't figure out how to account for the change in nucleus size.

Any ideas?

 

[voko]

What are the postulates of Bohr's model?

 

[dinospamoni]

Bohr's Model says that angular momentum is quantized, or only has certain values, so

$\vec{l} = \vec{r} \times \vec{p}$

simplified, it is

$l = rmv = n\hbar$

Right direction?

 

[voko]

That is not incorrect, but that does not seem to help much. What is the shape of orbits in Bohr's model, and what holds electrons in place?

 

[dinospamoni]

Ok so the electron has energy of -122.4 eV = -1.96*10^-17J

The total energy of the electron is:
$E = KE + U<br /> <br /> E = \frac{1}{2} m_e v^{2} + \frac{k q^2}{r}$

And by F = ma

$\frac{m_{e} v^{2}}{r} = \frac{k q^{2}}{r^2}$

I solved for v²

and found this expression for r:

$r = \frac{2 k q^{2}}{k q^{2}-2 E}$

However I found this radius to be 1.18 * 10^-11m which is wrong

 

[voko]

The equation you termed F = ma is not entirely correct, as you neglected the charge of the nucleus.

Then you could use the angular momentum rule.

 

[dinospamoni]

Ah I see.

so the force equation would be

$\frac{m_{e} v^{2}}{r} = \frac{k 6 q^{2}}{r^2}$?

 

[voko]

Ah I see.

so the force equation would be

$\frac{m_{e} v^{2}}{r} = \frac{k 6 q^{2}}{r^2}$?

Yes.