91影视

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 knoww that, energy of a photon

$E=\frac{hc}{位}$

Where, his Planck's constant, cis the speed of light, and$位$

wavelength of photon emitted.

Here, the numerical value of is given by,

$hc=1240eV.nm$

Define the energy at $450nm$as below.

$$\begin{gathered} \begin{array}{rcl}E& =& \frac{hc}{位}\\ & =& \frac{1240eV.nm}{450nm \\ =2.76eV \\ }\end{array} \end{gathered}$$

Define the energy at as below.

$$\begin{gathered} \begin{array}{rcl}E& =& \frac{hc}{位}\\ & =& \frac{1240eV.nm}{500nm \\ =2.49eV}\end{array} \end{gathered}$$

As you know that Energy of an electron in 鈥榥鈥� 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$4鈫�3.8鈫�4.9鈫�4$will correspond to energies$2.64eV$,$2.55eV$, and$2.73eV$which are in the given range.

$ifE=\frac{hc}{位}$

$$\begin{gathered} For4鈫�3 \\ 2.64eV=\frac{1240eV.nm}{位} \\ 位=470nm \end{gathered}$$

$$\begin{gathered} For8鈫�4 \\ 2.55eV=\frac{1240eV.nm}{位 \\ } \end{gathered}$$

$位=487nm$

role="math" localid="1659619345623" $$\begin{gathered} For9鈫�4 \\ 2.73eV=\frac{1240eV.nm}{位} \\ 位=455nm \end{gathered}$$

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