What is the energy equation in Schrodinger's Spherical equation?

[Danielk010]

Homework Statement

Show by direct substitution that the wave function corresponding to n = 1, l = 0, m l = 0 is a
solution of Eq. 7.10 corresponding to the ground-state energy of hydrogen.

Relevant Equations

Schrodinger's Spherical equation: $\frac {-\hbar^2} {2m} (\frac {d^2 \psi} {dr^2} + \frac {2} {r} * \frac {d\psi} {dr} + \frac {1} {r^2\sin(\theta)} * \frac {d} {d\theta} * \sin(\theta) *\frac {d\psi} {d\theta} + \frac {1} {r^2\sin(\theta)^2} * \frac {d^2\psi} {d\phi^2}) + U(r) * \psi (r, \theta, \phi) = E * \psi (r, \theta, \phi)$
Radial equation: $\frac {2*e^{\frac {-r} {a_0}}} {a_0^{\frac {3} {2}}}$
Theta equation: $\frac {1} {\sqrt 2}$
Phi equation: $\frac {1} {\sqrt {2\pi}}$
Wave function = Radial equation * Theta equation * Phi equation
Wave function: $\frac {e^{\frac {-r} {a_0}}} {\sqrt \pi * a_0^{\frac {3} {2}}}$
My final energy result: $\frac {2\pi\varepsilon_0 (-\hbar^2 * r + 2a_0^2\hbar^2) - ma_0^4e^2} {4\pi\varepsilon_0 r a_0^4 m}$

I attempted the problem by first finding the radial, theta, and phi equation for the ground state of a hydrogen atom. I multiplied the three equations to get the wave equation. From there, I took each derivative in the Schrodinger Spherical equation and found that $\frac {\partial^2 \psi} {\partial \phi^2} = 0$ and $\frac {\partial \psi} {\partial \theta} = 0$. From there I found that $\frac {\partial \psi} {\partial r} = \frac {-e^{\frac {-r} {a_0}}} {\sqrt \pi * a_0^{\frac {5} {2}}}$ and $\frac {\partial \psi^2} {\partial r^2} = \frac {e^{\frac {-r} {a_0}}} {\sqrt \pi * a_0^{\frac {7} {2}}}$. I then just plugged in the different equations including, $U(r) = \frac {-e^2} {4\pi\varepsilon_0 r}$ into the equation to get my final result for energy. Compared to $E_1 = \frac {-me^4} {32\pi^2\varepsilon_0^2\hbar^2}$ I am not sure if I did the math wrong to not get that energy equation or is there a different energy equation I should use. Thank you for any help provided.

 

[kuruman]

Radial equation: $\frac {2*e^{\frac {-r} {a_0}}} {a_0^{\frac {3} {2}}}$
Theta equation: $\frac {1} {\sqrt 2}$
Phi equation: $\frac {1} {\sqrt {2\pi}}$

These are not equations.

From there, I took each derivative in the Schrodinger Spherical equation and found that $\frac {\partial^2 \psi} {\partial \phi^2} = 0$ and $\frac {\partial \psi} {\partial \theta} = 0$. From there I found that $\frac {\partial \psi} {\partial r} = \frac {-e^{\frac {-r} {a_0}}} {\sqrt \pi * a_0^{\frac {5} {2}}}$ and $\frac {\partial \psi^2} {\partial r^2} = \frac {e^{\frac {-r} {a_0}}} {\sqrt \pi * a_0^{\frac {7} {2}}}$. I then just plugged in the different equations including, $U(r) = \frac {-e^2} {4\pi\varepsilon_0 r}$ into the equation to get my final result for energy. Compared to $E_1 = \frac {-me^4} {32\pi^2\varepsilon_0^2\hbar^2}$ I am not sure if I did the math wrong to not get that energy equation or is there a different energy equation I should use. Thank you for any help provided.

For the radial part you took the second derivative with respect to $r$ but it seems that you left out the $\frac{2}{r}\frac{\partial \psi}{\partial r}$ term.

 

[Danielk010]

These are not equations.


For the radial part you took the second derivative with respect to $r$ but it seems that you left out the $\frac{2}{r}\frac{\partial \psi}{\partial r}$ term.

I multiplied the $\frac{2}{r}$ after getting the derivative. Also what do you mean they are not equations?

 

[kuruman]

Also what do you mean they are not equations?

Equations usually have an "equals" sign that looks like $=$.
All I see is a colon that looks like $:$

After you take your derivatives and you get what you call your "final energy result", that's only half of an equation. Write an equation that separates two sides with an $=$ sign and has mathematical expressions on each side, not words on one and a mathematical expression on the other. If your "final energy result" is on the left-hand side, what should be on the right-hand side?

 

[Danielk010]

Equations usually have an "equals" sign that looks like $=$.
All I see is a colon that looks like $:$

After you take your derivatives and you get what you call your "final energy result", that's only half of an equation. Write an equation that separates two sides with an $=$ sign and has mathematical expressions on each side, not words on one and a mathematical expression on the other. If your "final energy result" is on the left-hand side, what should be on the right-hand side?

Oh my bad. For my final energy result it would E = *that equation*

 

[kuruman]

Oh my bad. For my final energy result it would E = *that equation*

You still have not written down an equation. Remember, mathematical expressions on both sides.

 

[Danielk010]

You still have not written down an equation. Remember, mathematical expressions on both sides.

Sorry. E = $\frac {2\pi\varepsilon_0 (-\hbar^2 * r + 2a_0^2\hbar^2) - ma_0^4e^2} {4\pi\varepsilon_0 r a_0^4 m}$

 

[kuruman]

Where did you get that E is equal to all those derivatives that you took?

 

[Danielk010]

Where did you get that E is equal to all those derivatives that you took?

I used the schordinger equation. I wrote the math down. Would it be better if I shared a picture of the math?

 

[kuruman]

I used the schordinger equation. I wrote the math down. Would it be better if I shared a picture of the math?

The right hand side of the equation is $E~\psi(r)$. Show me the equation that has your "final energy result" plus the potential energy on the left side and $E~\psi(r)$ on the right side. Yes, share a picture of the math but make sure it's legible.

 

[Danielk010]

The right hand side of the equation is $E~\psi(r)$. Show me the equation that has your "final energy result" plus the potential energy on the left side and $E~\psi(r)$ on the right side. Yes, share a picture of the math but make sure it's legible.

The $\psi(r)$ was canceled out.

 

[kuruman]

I deleted my immediate reply in case you saw it because I misread the $r$ in the denominator. I will post an more appropriate reply soon.

 

[Danielk010]

thank you so much for the help.

 

[kuruman]

Here is the more appropriate reply. First of all the powers of $a_0$ in the denominator are incorrect. If you divide $a_0^{7/2}$ by $a_0^{3/2}$, you don't get $a_0^4$ which is $a_0^{8/2}$. Same problem with $a_0^{5/2}$. Fix it.

In the corrected simplified equation (see below), replace ##U(r) with what it is equal to but do not add the fractions. Instead factor out the $\frac{1}{r}$ in the two terms that contain it and rewrite the equation.

Do you see where to go next? Ask if you don't.

 

[Danielk010]

Here is the more appropriate reply. First of all the powers of $a_0$ in the denominator are incorrect. If you divide $a_0^{7/2}$ by $a_0^{3/2}$, you don't get $a_0^4$ which is $a_0^{8/2}$. Same problem with $a_0^{5/2}$. Fix it.

In the corrected simplified equation (see below), replace ##U(r) with what it is equal to but do not add the fractions. Instead factor out the $\frac{1}{r}$ in the two terms that contain it and rewrite the equation.

View attachment 343474
Do you see where to go next? Ask if you don't.

So would this be correct?

 

[kuruman]

I said

In the corrected simplified equation (see below), replace ##U(r) with what it is equal to but do not add the fractions[/color]. Instead factor out the $\frac{1}{r}$ in the two terms that contain it and rewrite the equation.[/size]

Do it.

 

[Danielk010]

I said

Do it.

Sorry. Would this be correct?

 

[kuruman]

I works except that its sideways. Look at the equation. On the right you have the ground state state energy which is a constant. On the left you have a constant $C_1$ times of $\frac{1}{r}$ plus another constant $C_2$. How can that be reconciled? Hint: This equation must hold for any $r$.

Also, it looks like you are unsure about dividing powers. If the base is the same, you subtract the exponent in the denominator from the exponent in the numerator
$$\frac{a_0^{\frac{5}{2}}}{a_0^{\frac{3}{2}}}=a_0^{\frac{5}{2}-\frac{3}{2}}=a_0^{\frac{2}{2}}=a_0.$$Fix it before you proceed.

 

[Danielk010]

I works except that its sideways. Look at the equation. On the right you have the ground state state energy which is a constant. On the left you have a constant $C_1$ times of $\frac{1}{r}$ plus another constant $C_2$. How can that be reconciled? Hint: This equation must hold for any $r$.

Also, it looks like you are unsure about dividing powers. If the base is the same, you subtract the exponent in the denominator from the exponent in the numerator
$$\frac{a_0^{\frac{5}{2}}}{a_0^{\frac{3}{2}}}=a_0^{\frac{5}{2}-\frac{3}{2}}=a_0^{\frac{2}{2}}=a_0.$$Fix it before you proceed.

Thank you. Would equating r to be 1 work?

 

[kuruman]

Thank you. Would equating r to be 1 work?

Nope. Like I said, it should hold for any $r$, not just for the specific value of $r=1$. Besides, what does "1" mean? One meter. one inch, one light year, one what? You need to ponder this a bit that is why I am not giving you the answer. Just think. What must be true for E on the other side to be constant?

 

[Danielk010]

Nope. Like I said, it should hold for any $r$, not just for the specific value of $r=1$. Besides, what does "1" mean? One meter. one inch, one light year, one what? You need to ponder this a bit that is why I am not giving you the answer. Just think. What must be true for E on the other side to be constant?

Should you just plug in Bohr's radius?

 

[kuruman]

That would not work. Look, you have an equation that looks like
$E=\dfrac{1}{r}C_1+C_2$ where $C_1$ and $C_2$ are constants.
You want $E$ to be a constant. What must be true so that you can write
$E=const.$?
In other words, how can you get rid of the $\frac{1}{r}$ term?
Remember the equation is
$$\frac{1}{r}\left(\frac{\hbar^2}{ma_0}-\frac{e^2}{4\pi\epsilon_0}\right)-\frac{\hbar^2}{2ma_0^2}=E.$$

 

[Danielk010]

That would not work. Look, you have an equation that looks like
$E=\dfrac{1}{r}C_1+C_2$ where $C_1$ and $C_2$ are constants.
You want $E$ to be a constant. What must be true so that you can write
$E=const.$?
In other words, how can you get rid of the $\frac{1}{r}$ term?

Solve for r, and then plug it in? $r = \frac {C_1} {E-C_2}$

 

[kuruman]

No. Set $C_1=0.$ When you do that, which of the quantities in $C_1$ have fixed values?

 

[Danielk010]

No. Set $C_1=0.$ When you do that, which of the quantities in $C_1$ have fixed values?

all of the values except m

 

[kuruman]

Why except $m$? Isn't the mass of the electron fixed at $9.11\times 10^{-31}~$ kg?
What about the Bohr radius $a_0$? What is an expression for it, not a numerical value. Look it up if you don't remember.

 

[Danielk010]

Oh my bad. All of the values in Bohr's radius are constants if I am not mistaken. $a_0 = \frac {4\pi\varepsilon_0\hbar^2} {m_e*e}$. 4, \pi, \hbar and the permittivity are all constants. An electron's rest mass is constant ($0.511 Mev$), and the elementary charge of an electron is also a constant.

 

[kuruman]

What happens when you substitute the expression that you quoted for $a_0$ in $$\frac{1}{r}\left(\frac{\hbar^2}{ma_0}-\frac{e^2}{4\pi\epsilon_0}\right)-\frac{\hbar^2}{2ma_0^2}=E~~?$$

 

[Danielk010]

What happens when you substitute the expression that you quoted for $a_0$ in $$\frac{1}{r}\left(\frac{\hbar^2}{ma_0}-\frac{e^2}{4\pi\epsilon_0}\right)-\frac{\hbar^2}{2ma_0^2}=E~~?$$

oohhh. I got it. thank you for the help

 

[kuruman]

The expression for the Bohr radius is obtained by setting the coefficient of the $\dfrac{1}{r}$ term equal to zero. This ensures that the energy of the ground state is constant for any value of variable $r$.