Ground state energy for hydrogen-like “atom”

[Andy123]

Homework Statement

For a hydrogen-like “atom” (e.g., He+ ion), with nuclear charge Ze, it is claimed that the the ground-state wavefunction is spherically symmetric and is given by ψ(r)=Aexp(−αr) , where A and α are constant. (a) Determine the normalization constant A in terms of α. (b) Determine α by making use of (Eq.3) in the lecture note (Ch.9 pg 4). Note that you need to make some change in the equation before applying it to this system. Hence, find the ground state energy.

The mentioned equation here is attached below. The R in the equation refers to ψ(r) and l is the orbital quantum number[/B]

Homework Equations

The Attempt at a Solution

I can finish part a) but for part b), i substitute ψ(r)=Aexp(−αr) and try to solve for α. It comes up with the following result and I am stuck. please help![/B]

 

[mfb]

It is hard to read your handwriting, can you write it here (as text or with LaTeX)?
You know l for the ground state, that allows to simplify the equation significantly.

 

[TSny]

Hi, Andy. Welcome to PF!

Looks like you dropped a factor of $\hbar$ in going from the first to second line.

Look at your 6th line (before you solved for $\alpha$). Note that it must be satisfied for all values of $r$ even though $\alpha$ is a constant.

 

[Andy123]

It is hard to read your handwriting, can you write it here (as text or with LaTeX)?
You know l for the ground state, that allows to simplify the equation significantly.

I'm sorry my hand-writing is a bit messy. I will try to learn how to use LaTex

 

[Andy123]

Hi, Andy. Welcome to PF!

Looks like you dropped a factor of $\hbar$ in going from the first to second line.

Look at your 6th line (before you solved for $\alpha$). Note that it must be satisfied for all values of $r$ even though $\alpha$ is a constant.

I understand that $\alpha$ is a constant and when i solve it, it somehow seems like to depend on r. And when i check my calculations, it seems to be ok except that $\hbar$ part. What shall i do? Can i just randomly let r be any number i want?

 

[mfb]

For the correct α, the equation does not depend on r any more. Or, in other words, the derivative of "something=0" with respect to r will be zero.

 

[TSny]

I understand that $\alpha$ is a constant and when i solve it, it somehow seems like to depend on r. And when i check my calculations, it seems to be ok except that $\hbar$ part. What shall i do? Can i just randomly let r be any number i want?

Right. $\alpha$ can't depend on $r$. This fact actually determines the value of $\alpha$. By examining your 6th equation, can you see a way for the two terms that depend on $r$ to cancel out?

 

[Andy123]

Right. $\alpha$ can't depend on $r$. This fact actually determines the value of $\alpha$. By examining your 6th equation, can you see a way for the two terms that depend on $r$ to cancel out?

since $\alpha$. doesn't depends on $r$ and $E$ doesn't depend on $r$ too, the only way to cancel them is to put those two terms to be zero when added up and then i can solve for $\alpha$.?

 

[TSny]

Yes, I think that's right. Combine the two terms that contain $r$ and see if you can make the combined term vanish.

 

[Andy123]

Yes. I can find the answer now. Thanks all of you!