Question

The Shrodinger wave equation for hydrogen atom is
\[\huge{\psi_{2s}=\frac{1}{4(2 \pi )^\frac{1}{2}}(\frac{1}{a_0})^\frac{3}{2}(2-\frac{r_0}{a_0})e^{-r/2a_0}}\]where \[a_0=\text{Bohr's radius}.\]Let the radial node in 2s be at \[\LARGE{r_0}.\]Then find \[\LARGE{r_0}\] in terms of \[\LARGE {a_0}.\]

Answer 1

Its solution is as follows but couldn't understood it.
This solution is given in my book.

Solution: -
At radial node; \[\psi^2 \space \space \text{must vanish.}\]Therefore; \[\LARGE{\psi_{2s}^2=0=[\frac{1}{4\sqrt{2 \pi}}]^2[2-\frac{r_0}{a_0}]^2e^{-\frac{r}{a_0}}.}\]\[\implies 2-\frac{r_0}{a_0}=0.\]\[\implies r_0=2a_0.\]So please help me understanding this:)

Answer 2

The only thing I can imagine that might be causing you trouble is figuring out why you don't have to worry about the e^(-r/ao) bit. The reason is that you have three factors on the right hand side, each multiplying each other. If any ONE of them equals zero, then the entire RHS equals zero.

Now take the three factors one at a time. The first is a numerical constant, and can never be zero. The third, the exponential, is also never zero. (Get your calculator to plot e^(-x) to see this for yourself.) That leaves only the second factor: if and only if it's zero will the RHS be zero. So that brings you to the second equation in your solution, where you set the second factor equal to zero and solve for what value of r0 makes the equation true.

Answer 3

But wt about
\[\LARGE{\color{blue}{\frac{1}{a_0}.}}\] ???????????????????????????

Answer 4

@Carl_Pham

Answer 5

Why in the solution \[1 \over a_0\] is considered?????????????

Answer 6

ao is given, and you have to solve for ro.

Plug in ro = 1/ao in your equation, and you will see that psi² is not zero.

Besides, ro must be a distance, whereas 1/ao is not.

Answer 7

oh ok
thanx:)