The Rydberg equation (1/lambda=R/ni^2–R/nf^2) can be treated as a line equation. What is the value of nf as a function of the slope (m) and y-intercept(b)?
by GC 1/lambda = R/ni^2–R/nf^2
y = mx + b (standard form) of a linear equation)
x = (y-b)/m
Let,
y = 1/lambda
m = -R
x = 1/nf^2
b = R/ni^2
1/nf^2 = [(1/lamda)-(R/ni^2)]/(-R)
Solve for nf
i still don't get how you get rid of y and x to get m and b in the end
Answer is (m/b)^1/2
I don't know the answer to this question but I put the answer given above is wrong if you are doing the uci webwork from chem 1LB
dude its (-M/B)^(1/2)
im doing uci webwork and that's right
(-M/B)^(1/2)
Alex Le and bowisha
you're all probably stupid in school
thats y u want to let people down idiots
answer is correct as mb to the 1/2
not negative
this is the Right answer
I'm doing the webwork for uci, not sure what the answer is but it's not (m/b)^-1/2. thanks people....
it's negative
yo it is (-m/b)^(1/2) i just did it and submitted.
To get rid of y and x in the Rydberg equation and express nf as a function of the slope (m) and y-intercept (b), you need to rearrange the equation to match the form of a linear equation (y = mx + b). Here's how you can do it:
1. Start with the Rydberg equation: 1/lambda = R/ni^2 – R/nf^2.
2. Substitute the variables as follows:
- Let y = 1/lambda
- Let m = -R
- Let x = 1/nf^2
- Let b = R/ni^2
3. Now rewrite the Rydberg equation with the substitutions:
1/lambda = m/ni^2 – m/nf^2
4. Rearrange the equation to isolate nf:
1/lambda = m/ni^2 – m/nf^2
Multiply both sides by nf^2 to remove the fraction:
nf^2/lambda = m/ni^2 – m
Add m to both sides:
nf^2/lambda + m = m/ni^2
Subtract m/ni^2 from both sides:
nf^2/lambda = -m/ni^2
Multiply both sides by lambda to isolate nf^2:
nf^2 = -lambda*m/ni^2
5. Take the square root of both sides to solve for nf:
nf = √(-lambda*m/ni^2)
Therefore, nf is a function of the slope (m) and y-intercept (b) and can be expressed as nf = √(-lambda*m/ni^2).