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

• Danielk010
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.

Danielk010 said:
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.

Danielk010 said:
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.

kuruman said:
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?

Danielk010 said:
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?

kuruman said:
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*

Danielk010 said:
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.

kuruman said:
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} ##

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

kuruman said:
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?

Danielk010 said:
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.

kuruman said:
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.

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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.

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.

kuruman said:
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?

I said
kuruman said:
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 it.

kuruman said:
I said

Do it.
Sorry. Would this be correct?

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.

kuruman said:
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?

Danielk010 said:
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?

kuruman said:
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?

Last edited:
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
$$\frac{1}{r}\left(\frac{\hbar^2}{ma_0}-\frac{e^2}{4\pi\epsilon_0}\right)-\frac{\hbar^2}{2ma_0^2}=E.$$

Last edited:
kuruman said:
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} ##

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

kuruman said:
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

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.

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.

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

kuruman said:
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

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##.

Danielk010

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