91Ó°ÊÓ

When an electron transitions from a higher orbit to the lower orbit, it loses energy in the form of photons. And the energy of that photon is equal to the difference in energy of the transition states in which the transition is observed.

As you know that, energy of a photon
$E=\frac{hc}{\mathrm{λ}}$

Where, his Planck's constant, cis the speed of light, and localid="1659382795937" $\mathrm{λ}$wavelength of photon emitted.

Here, the numerical value of is given by,

$hc=1240eV.nm$

Define the energy atlocalid="1659618186089" $450nm$ as below.
$$\begin{gathered} E=\frac{hc}{\mathrm{λ}} \\ =\frac{1240eV.nm}{450nm \\ =2.76eV \\ } \end{gathered}$$

Define the energy at$450$ as below.

$$\begin{gathered} E=\frac{hc}{\mathrm{λ}} \\ =\frac{1240eV.nm}{500nm \\ =2.49eV} \end{gathered}$$

As you know that Energy of an electron in ‘n’ th orbit is given by,
$E=-z^2\frac{13.6eV}{n^2}(n=1,2,3............)$

Where, $z$is the atomic number of hydrogen-like atom and$n$ is the principal quantum number.

$$\begin{gathered} E_n=-2^2\frac{13.6eV}{n^2 \\ } \end{gathered}$$
$E_n=-\frac{54.4eV}{n^2}$ .......(1)
Now, using equation (1), you get.

$$\begin{gathered} E_1=-54.4eV,E_2=-13.6eV,E_3=6.04eV,E_4=3.4eV,E_5=2.18eV,E_6=1.51eV. \\ {E}_7=1.11eV,E_8=0.85eV,E_9=0.67eV,E_{10}=0.54eV,... \end{gathered}$$

From step 2, you get,

The transitions$4→3.8→4.9→4$will correspond to energieslocalid="1659618569980" $2.64eV$, $2.55eV$and $2.73eV$which are in the given range.

$ifE=\frac{hc}{\mathrm{λ}}$

$$\begin{gathered} For4→3 \\ 2.64eV=\frac{1240eV.nm}{\mathrm{λ}} \\ \mathrm{λ}=470nm \end{gathered}$$

localid="1659381996658" $$\begin{gathered} For8→4 \\ 2.55eV=\frac{1240eV.nm}{\mathrm{λ} \\ } \end{gathered}$$
$\mathrm{λ}=487nm$




$$\begin{gathered} For9→4 \\ 2.73eV=\frac{1240eV.nm}{\mathrm{λ}} \\ \mathrm{λ}=455nm \end{gathered}$$

Wavelengths of the photons corresponding to transitions $4→3.8→4.9→4$in the given range are$470nm$, 487nm and 455 nm respectively.