Question

My algebra is a little rusty. Given the following equation:\[\tilde{\nu}=\frac{1}{\lambda}=R\bigg(\frac{Z^2}{n^2_i}-\frac{Z^2}{n^2_i}\bigg)\]
Is there a better way to solve for λ than this?\[\lambda=\frac{1}{\nu}=\frac{1}{R(Z^2/n^2_i-Z^2/n^2_f)}\]
It seems like I should be able to have everything on the either the top or bottom of the fraction, instead of embedding fractions.

Answer 1

You are essentially raising the R() part to the -1st power. Powers don't distribute, so there isn't an easy way to not have the embedded fractions.

Answer 2

* powers don't distribute across addition or subtraction

Answer 3

There is no way to take the following terms and do something graceful with them?\[.../n^2_iand.../n^2_f\]

Answer 4

not to my knowlege, but I could be wrong.

Answer 5

Thank you much.

Answer 6

I plugged this problem into WolframAlpha, and it gave me an idea for simplifying it. Does this work?\[\lambda = \frac{1}{\tilde \nu}=-\frac{n_i^2n_f^2}{RZ^2(n_i^2-n_f^2)} = -\bigg(\frac{1}{R(n_i^2-n_f^2)}\bigg)\bigg(\frac{n_in_f}{Z}\bigg)^2\]

Answer 7

yes its right

Answer 8

this the equation comes from Bhor's model..

Answer 9

i think the first equation is easiest to deal with because for an atom u know the Z and Rhydbarg's constant u know , it is 109737 cm^-1 and if u put only the value of ni and nf then its very easy to calculate..
and then if u want to find wave no. then just take the reciprocal of the wave length..
this approach reduces the effort...
its my personal opinion ,as far as i have deal with this kind of things.

Answer 10

So, presumably I'm looking at something like this for the most compact form of the formula?\[λ=\tilde ν^{-1}=(RZ^2(n^{-2}_i−n^{-2}_i))^{-1}\]

Answer 11

yes..

Answer 12

Thank you much.

Answer 13

you most welcome...