91影视

The given data is listed below.

Wavelength of light is $\mathrm{位}=121.6\mathrm{nm}$.

The principal quantum number is used to describe the electron鈥檚 state and is the one-four quantum number assigned to each electron in an atom.

The value of the principal quantum number is a natural number.

The energy emitted by the hydrogen atom of the photon is expressed as,

$E_{ph}=A\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Now, multiply both sides by $\frac{1}{hc}$, and you have

$\frac{E_{ph}}{hc}=\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Also as the wavelength is,

$\frac{1}{\mathrm{位}}=\frac{{\mathrm{E}}_{\mathrm{Ph}}}{\mathrm{hc}}$

Substitute$\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$for$\frac{{\mathrm{E}}_{\mathrm{P}}\mathrm{h}}{\mathrm{hc}}$in the above equation.

$$\begin{gathered} \frac{1}{位}=\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\ \frac{hc}{位A}=\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \end{gathered}$$

Substitute 121.6 nm for $位$, $6.26脳{10}^{-34}\mathrm{J}.\mathrm{s}$for h , $3脳{10}^8\mathrm{m}/\mathrm{s}$for c , and 13.6 eV for A in the above equation.

$$\begin{gathered} \frac{\left(6.26脳{10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3脳{10}^8\mathrm{m}/\mathrm{s}\right)}{\left(121.6脳{10}^{-9}\mathrm{m}\right)\left(13.6脳1.602脳{10}^{-9}\mathrm{J}\right)}=\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right) \\ 0.75=\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right) \end{gathered}$$

To satisfy the above equation, the values of ${\mathrm{n}}_2$and ${\mathrm{n}}_1$are 1 and 2 respectively. Thus,

$\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}=0.75$

Thus, the higher quantum number is ${\mathrm{n}}_1=2$.

And so,

$\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}=0.75$

Thus, the lower quantum number is $n_2=1$.

The name of the series that includes the transition is Lyman series transition as this transition occurs when the electron goes from $n_1$to $n_2$emitting a photon.