Coincident spectral lines. ${}^{\mathbf{43}}$According to the Rydberg formula (Equation 4.93) the wavelength of a line in the hydrogen spectrum is determined by the principal quantum numbers of the initial and final states. Find two distinct pairs$\left\{n_i,n_f\right\}$ that yield the same $\mathit{\lambda }$. For example,role="math" localid="1656311200820" $\left\{6851,6409\right\}$ and$\left\{15283,11687\right\}$will do it, but you're not allowed to use those!

The Rydberg formula is a mathematical method for calculating light wavelength. The energy of an electron varies when it moves from one atomic orbit to another.

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_{\mathrm{f}}^2}‐\frac{1}{{\mathrm{n}}_{\mathrm{i}}^2}\right)$

Out mission in this problem is to find the four numbers such that they produce the same spectral lines.

According to the Rydberg formula:

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left|\left(\frac{\mathit{1}}{{\mathit{n}}_{\mathit{f}}^{\mathit{2}}}\mathit{‐}\frac{\mathit{1}}{{\mathit{n}}_{\mathit{i}}^{\mathit{2}}}\right)\right|$

It is very difficult to just guess some values.

The given numbers to find new numbers, the given pair of numbers are $\left[6851,6409\right]$and $\left[15283,11687\right]$.

For these two pairs, find their prime factors:

$$\begin{gathered} 6851=13\times 17\times 31 \\ 6409=13\times 17\times 29 \\ 15283=17\times 29\times 31 \\ 11687=13\times 29\times 31 \end{gathered}$$

The first pair of numbers have two common factors and also the second pair have two common factors, where each member of the first pair uses one of the prime factors of the second pair, and vice versa.

The numbers can be written as:

$$\begin{gathered} \frac{1}{{\left(6409\right)}^2}-\frac{1}{{\left(6851\right)}^2}=\frac{1}{{\left(11687\right)}^2}-\frac{1}{{\left(15283\right)}^2} \\ \frac{1}{{\left(13\times 17\times 29\right)}^2}-\frac{1}{{\left(13\times 17\times 31\right)}^2}=\frac{1}{{\left(13\times 29\times 31\right)}^2}-\frac{1}{{\left(17\times 29\times 31\right)}^2} \\ \mathrm{Let}\mathrm{a}=31,\mathrm{b}=29,\mathrm{c}=17\mathrm{and}\mathrm{d}=13,\mathrm{then}\mathrm{we}\mathrm{can}\mathrm{write}: \\ \frac{1}{{\left(bcd\right)}^2}-\frac{1}{{\left(acd\right)}^2}=\frac{1}{{\left(abd\right)}^2}-\frac{1}{{\left(abc\right)}^2} \\ \frac{a^2-b^2}{{\left(abcd\right)}^2}=\frac{c^2-d^2}{{\left(abcd\right)}^2} \\ {a}^2-b^2=c^2-d^2 \\ \left(a+b\right)\left(a-b\right)=\left(c+d\right)\left(c-d\right) \end{gathered}$$

The method is to choose two prime numbers a and b, then we find the difference $a^2-b^2$after that we find the factors of the difference, from these factors we take two number and such that$\left(c+d\right)\left(c-d\right)=a^2-b^2$

Let $a=7$and $b=5$then $a^2-b^2=24$which has factors of 1,2,3, 6,8 and 12 note that we have already used 2 and 12

Therefore $a+b=12$and $a-b=2$we pick another two numbers, say, 4 and 6 , that is:

$$\begin{gathered} c=5 \\ c+d=6 \\ 5+d=6 \\ d=6-5 \\ d=1 \end{gathered}$$

Substitute $d=1$in the above equation

$$\begin{gathered} c-d=4 \\ c-1=4 \\ c=4+1 \\ c=5 \end{gathered}$$

But :

$\frac{1}{{\left(bcd\right)}^2}-\frac{1}{{\left(acd\right)}^2}=\frac{1}{{\left(abd\right)}^2}-\frac{1}{{\left(abc\right)}^2}$

where the pairs are $\left[bcd,acd\right]$and $\left[abd,abc\right]$,so:

$\frac{1}{{25}^2}-\frac{1}{{35}^2}=\frac{1}{{35}^2}-\frac{1}{{175}^2}$

So the pairs are:

$\left[35,25\right]\left[175,35\right]$

Follow this method to find another pairs but you have to be careful, since not all the numbers work using this method, so you have to check your pairs.

Therefore, the two distinct pairs $\left\{n_i,n_f\right\}$that yield the same $\lambda$is$\left[35,25\right]\left[175,35\right]$