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)
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.
To rearrange Balmer's equation to solve for n2, you'll need to manipulate the equation using algebraic principles. Here's a step-by-step process to do that:
1. Start with the equation:
1/λ = R(1/2^2 - 1/n2^2)
2. Begin by simplifying the terms inside the parentheses by evaluating the exponents:
1/λ = R(1/4 - 1/n2^2)
3. Next, find a common denominator for 1/4 and 1/n2^2, which is 4n2^2:
1/λ = R((4 - 4/n2^2)/4n2^2)
4. Simplify the numerator by distributing the factor of 4:
1/λ = R((4n2^2 - 4)/4n2^2)
5. Combine like terms in the numerator:
1/λ = R(4(n2^2 - 1)/4n2^2)
6. Cancel out the common factors of 4 in the numerator and denominator:
1/λ = R(n2^2 - 1)/n2^2
7. Invert both sides of the equation to isolate n2:
λ = n2^2/(n2^2 - 1)
Now you have the rearranged equation! To solve for n2, you can substitute the known value of λ into the equation and solve for n2 using algebraic operations.