I'm unsure of how to rearrange Balmer's equation to solve for n2. The equation is 1/lambda= R(1/2^2-1/n2^2)
3 answers
If you have a problem with numbers, the easy thing to do is to leave the equation as is, substitute the numbers, the solve for the unknown in the equation. If you get stuck, post the numbers and someone can help you through.
I don't have to plug on any numbers I just have to get n2 by itself, but I don't know how to isolate it.
Let's let lambda be L and drop the n1 and n2. Since n1 = then that squared is 1/4 and n2^2 will just be n^2
1/L = R(1/4 - 1/n^2)
1 = RL(0.25-1/n^2)
1 = RL[(0.25n^2-1)/n^2]
n^2 = RL(0.25n^2-1)
n^2 = 0.25RLn^2-RL
n^2-0.25RLn^2 = RL
n^2(1-0.25RL) = RL
n^2 = [RL/(1-0.25RL)]
and take the square root of that. Check my algebra. I slip more signs and drop more coefficients than you can imagine.
1/L = R(1/4 - 1/n^2)
1 = RL(0.25-1/n^2)
1 = RL[(0.25n^2-1)/n^2]
n^2 = RL(0.25n^2-1)
n^2 = 0.25RLn^2-RL
n^2-0.25RLn^2 = RL
n^2(1-0.25RL) = RL
n^2 = [RL/(1-0.25RL)]
and take the square root of that. Check my algebra. I slip more signs and drop more coefficients than you can imagine.
1 answer