Deriving the ground state energy of a hydrogen atom?

[21joanna12]

Homework Statement

Hello! I am trying to derive the ground state enegry of a hydrogen atom, and have come to

$U=\frac{-mk_{0}^{2}Ze^{4}}{n^{2}\hbar^{2}}$

Problem is, I know there should e another factor of 2 in the denomenator because I get the ground state energy of hydrogen as being 27.145eV rather than 13.6eV

Homework Equations

Centripetal force is $F_{c}=\frac{mv^{2}}{r}$

Coulomb's law is $F=\frac{k_{0}q_{1}q_{2}}{r^{2}}$

Potential is $U=\frac{k_{0}q_{1}q_{2}}{r}$

And using Bohr's idea that the angular momentum can only be multiplies of $\hbar$, so

$mvr=n\hbar$

so $v=\frac{n\hbar}{mr}$

$v^{2}=\frac{n^{2}\hbar^{2}}{m^{2}r^{2}}$

The Attempt at a Solution

For a system with a single electron orbiting a nucleus with Z protons, and equating the centripetal force with Coulomb force,

$\frac{mv^{2}}{r}=\frac{k_{0}Ze^{2}}{r^{2}}$

$mv^{2}=\frac{k_{0}Ze^{2}}{r}$

Now substituting in for $v^{2}$,

$\frac{n^{2}\hbar^{2}}{m^{2}r^{2}}=\frac{k_{0}Ze^{2}}{mr}$

So $r=\frac{n^{2}\hbar^{2}}{mke^{2}}$

Substituting this is for the potential,

$U=\frac{-mk_{0}^{2}Ze^{4}}{n^{2}\hbar^{2}}$

Please tell me where have gone wrong!
Thank you in advance :)

 

[throneoo]

13.6eV is the total energy of that system...so in addition to potential energy , you'll have to take kinetic energy into consideration , which you apparently have not

 

[21joanna12]

13.6eV is the total energy of that system...so in addition to potential energy , you'll have to take kinetic energy into consideration , which you apparently have not

Hi throneoo, thanks for your reply!

How would I take into account the kinetic energy to derive the 13.6eV? Also, I'm not too sure that kinetic energy is the problem here because the answer I got was exactly twice what it should be, so I think it may be more likely that I missed a factor of a half somewhere. Though I am not too sure...

 

[Simon Bridge]

You find the kinetic energy and add it to the potential energy.
You did make a miscalculation though ... what sign should the potential energy be?

 

[Orodruin]

How would I take into account the kinetic energy to derive the 13.6eV?

mv2=k0Ze2r

Does $mv^2$ remind you of something that you would typically see in an expression for kinetic energy? This would be the hand-waving approach but would be similar in rigour to applying the Bohr model. Alternatively you could actually compute the ground state wave function and its corresponding kinetic energy term - or apply the virial theorem.

 

[throneoo]

mv2=k0Ze2rmv^{2}=\frac{k_{0}Ze^{2}}{r}

you are very close to deriving the expression for the kinetic energy in terms of the potential energy...just derive it and plug it into the expressionfor the total energy . that should you a ground state energy of -13.6eV