91Ó°ÊÓ

Let the given released by a hydrogen atom in its n=4 state

The goal is to figure out all of the photon wavelengths that can be emitted from the hydrogen atom's fourth energy level.

We must first comprehend the mechanism that causes the emission before we can solve the problem.

The hydrogen atom is a quantized system, which implies it has discrete rather than continuous energy levels.

when an electron takes a quantum leap from a higher energy level $\left({\mathrm{E}}_{\mathrm{i}}\right)$to a lower energy level role="math" localid="1650734041991" $\left({\mathrm{E}}_{\mathrm{j}}\right)$It emits a photon when the following conditions are met.

From the ${\mathrm{i}}^{\mathrm{th}}$to the ${\mathrm{j}}^{\mathrm{th}}$level, the energy of the emitted photon,

$\left({\mathrm{E}}_{\mathrm{i}→\mathrm{j}}\right)={\mathrm{Δ·¡}}_{\mathrm{ij}}$

$={\mathrm{E}}_{\mathrm{i}}−{\mathrm{E}}_{\mathrm{j}}$

where ${\mathrm{E}}_{\mathrm{i}}$denotes the higher energy level (i) and${\mathrm{E}}_{\mathrm{j}}$denotes the lower energy level (j).

As a result, from the 4th state in the hydrogen atom, all potential photon energies are given as follows:

${\mathrm{E}}_{4→\mathrm{j}}={\mathrm{E}}_4−{\mathrm{E}}_{\mathrm{j}}\text{(where}\mathrm{j}=1,2,3)…\left(\mathrm{i}\right)$

The energy of a particular state n for the hydrogen atom where Z=1 is calculated using Eq.(38.38).

${\mathrm{E}}_{\mathrm{n}}=\frac{−13.6\mathrm{eV}}{{\mathrm{n}}^2}…\left(2\right)$

We get (2) by swapping it for (1).

$$\begin{gathered} {\mathrm{E}}_{4→\mathrm{n}}=\frac{−13.6\mathrm{eV}}{4^2}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}\text{(where}\mathrm{n}=1,2,3\text{)} \\ =\frac{\mathrm{hc}}{{\mathrm{λ}}_{4→\mathrm{n}}}, \end{gathered}$$

As result,$\boxed{{\mathrm{λ}}_{4→\mathrm{n}}=\frac{\mathrm{hc}}{\left(\frac{−13.6\mathrm{eV}}{4^2}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}\right)}\text{(where}\mathrm{n}=1,2,3\text{)}}$

There was a decrease in the number of participants n = 4 to n = 3

$$\begin{gathered} {\mathrm{λ}}_{4→3}=\frac{4.1357×{10}^{−15}\cancel{\mathrm{eV}}.\cancel{\mathrm{s}}×3×{10}^8â‹\dots \cancel{{\mathrm{m}.\mathrm{s}}^{−1}}}{\left(\frac{−13.6\cancel{\mathrm{eV}}}{4^2}+\frac{13.6\cancel{\mathrm{eV}}}{3^2}\right)} \\ ≃1.877×{10}^{−6}\mathrm{m}=\boxed{1877\mathrm{nm}} \end{gathered}$$

There was a decrease in the number of participants n = 4 to n = 2

$$\begin{gathered} {\mathrm{λ}}_{4→2}=\frac{4.1357×{10}^{−15}\cancel{\mathrm{eV}}.\cancel{\mathrm{s}}×3×{10}^8â‹\dots \cancel{{\mathrm{m}.\mathrm{s}}^{−1}}}{\left(\frac{−13.6\cancel{\mathrm{eV}}}{4^2}+\frac{13.6\cancel{\mathrm{eV}}}{2^2}\right)} \\ ≃4.866×{10}^{−7}\mathrm{m}=\boxed{486.6\mathrm{nm}} \end{gathered}$$

There was a decrease in the number of participants n = 4 to n = 1

$$\begin{gathered} {\mathrm{λ}}_{4→1}=\frac{4.1357×{10}^{−15}\cancel{\mathrm{eV}}.\cancel{\mathrm{s}}×3×{10}^8â‹\dots \cancel{{\mathrm{m}.\mathrm{s}}^{−1}}}{\left(\frac{−13.6\cancel{\mathrm{eV}}}{4^2}+\frac{13.6\cancel{\mathrm{eV}}}{1^2}\right)} \\ =9.73×{10}^{−8}\mathrm{m}=\boxed{97.3\mathrm{nm}} \end{gathered}$$

Three different wavelengths are possible$1877\mathrm{nm},486.6\mathrm{nm}\text{and}97.3\mathrm{nm}\text{.}$