- #1

perplexabot

Gold Member

- 329

- 5

Hey all. I have posted a thread regarding this question a while back. I did get an answer and everything. (Here is my old post along with the original question if you are interested: https://www.physicsforums.com/showthread.php?t=592885).

So i tried doing that problem again like this:

Given:

L = Hbar * √(l*(l+1))

Binomial expansion approximation: (1+z)^n = 1 + nz

My try:

L = Hbar * √l * (1 + l/2)

If you opened the link of my previous post you will see that I must obtain L = Hbar * (l + 1/2). As you can see my just calculated expression is not the same as the previous expression, however plugging in a certain value of l, one obtains the same value for either expression which leads me to believe the two expressions are equivalent. So my question is: How would I play with my current equation to obtain the equation from the previous post. Yes, this is completely an algebra question. Thank you.

So i tried doing that problem again like this:

Given:

L = Hbar * √(l*(l+1))

Binomial expansion approximation: (1+z)^n = 1 + nz

My try:

L = Hbar * √l * (1 + l/2)

If you opened the link of my previous post you will see that I must obtain L = Hbar * (l + 1/2). As you can see my just calculated expression is not the same as the previous expression, however plugging in a certain value of l, one obtains the same value for either expression which leads me to believe the two expressions are equivalent. So my question is: How would I play with my current equation to obtain the equation from the previous post. Yes, this is completely an algebra question. Thank you.

Last edited: