Distance of electron from nucleus

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SUMMARY

The discussion focuses on calculating the distance of an electron from the nucleus in a hydrogenic carbon-12 atom using Bohr's model. The relevant equation for the radius in excited states is given by r_n = n^2 × a_0, where a_0 is the Bohr radius. The participant struggles with adjusting the calculations for the carbon nucleus size and correctly applying the force equations. The correct approach involves considering the nucleus's charge and using angular momentum quantization principles to derive the electron's radius accurately.

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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?
 
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What are the postulates of Bohr's model?
 
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?
 
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?
 
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
 
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.
 
Ah I see.

so the force equation would be

\frac{m_{e} v^{2}}{r} = \frac{k 6 q^{2}}{r^2}?
 
dinospamoni said:
Ah I see.

so the force equation would be

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

Yes.
 

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