Question

What lie the winding of the Irasuliens? $3\to 2.4\to 2$, and $5\to 2$in the hydronics ${\mathrm{O}}^{-7}$? In what spectral range do these lie?

Short answer

${\lambda }_{32}=10.26\mathrm{nm};{\lambda }_{42}=7.6\mathrm{nm};{\lambda }_{52}=6.79\mathrm{nm}$

Step-by-step solution

Step: 1 Given information

The given hydric lie are${\mathrm{O}}^{-7}$

Step: 2 Calculation

We can begin by expressing our energy levels as follows:

$$\begin{gathered} E_n=-\frac{13.6Z^2\left(\mathrm{eV}\right)}{n^2} \\ {E}_3-E_2=-13.6Z^2\left(\mathrm{eV}\right)·\left(\frac{1}{3^2}-\frac{1}{2^2}\right) \end{gathered}$$$$\begin{gathered} E_3-E_2=\frac{hc}{{\lambda }_{32}} \\ {\lambda }_{32}=\frac{hc}{E_3-E_2} \\ {\lambda }_{32}=\frac{6.63·{10}^{-34}·3·{10}^8}{-13.6·8^2·1.6·{10}^{-19}·\left(\frac{1}{3^2}-\frac{1}{2^2}\right)} \\ {\lambda }_{32}=10.26\mathrm{nm} \end{gathered}$$

In the same approach, we can determine out $4\to 2$and$5\to 2$ transitions:

$$\begin{gathered} {\lambda }_{42}=\frac{hc}{E_4-E_2} \\ {\lambda }_{42}=\frac{6.63·{10}^{-34}·3·{10}^8}{-13.6·8^2·1.6·{10}^{-19}·\left(\frac{1}{4^2}-\frac{1}{2^2}\right)} \end{gathered}$$

$$\begin{gathered} {\lambda }_{42}=7.6\mathrm{nm} \\ {\lambda }_{52}=\frac{hc}{E_5-E_2} \\ {\lambda }_{52}=\frac{6.63·{10}^{-34}·3·{10}^8}{-13.6·8^2·1.6·{10}^{-19}·\left(\frac{1}{5^2}-\frac{1}{2^2}\right)} \\ {\lambda }_{52}=6.79\mathrm{nm} \end{gathered}$$

These wavelengths are all in the UV range.