Respuesta :
Solution :
Given that the violet line from Balmer series having a recorded wavelength of 410 nm. As the recorded wavelength is 410 nm we know, the wavelength range of violet light is 380 nm -450 nm. So the 410 nm is corresponding to violet light wavelength.
We know, among all the series of lines spectrum of [tex]$H_2$[/tex] atom, only the Balmer series are in visible range. For Balmer series, the equation of wavelength associated with the e-transition of n = 2 orbit to n > 2 is
[tex]$\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$[/tex]
where, R = Rydberg constant = 109678 per cm
λ = wavelength
Here the recorded λ = 410 nm = [tex]$410 \times 14^{-7} \ cm$[/tex]
∴ [tex]$\frac{1}{410 \times 10^{-7}}=109678\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$[/tex]
[tex]$\frac{1}{2^2}-\frac{1}{n^2}=\frac{1}{410 \times 10^{-7}\times 109678}$[/tex]
[tex]$\frac{1}{4}-\frac{1}{n^2}=0.222$[/tex]
[tex]$\frac{1}{n^2}=\frac{1}{4}-0.222$[/tex]
[tex]$n^2=\frac{1}{0.0276}$[/tex]
[tex]$n^2=36.2$[/tex]
[tex]$n^2 $[/tex] ≈ 36
n = ±6
Here, n is the principle quantum number. It cannot be negative, so the value of n = 6 for hydrogen atom if the violet line from Balmer series that has a recorded wavelength of 410 nm.
The value of n for the Hydrogen transition in the Balmer series with the wavelength 410 nm has been 6.
The principal quantum number in the Balmer series transition has been given by n. The minimum transition for the Balmer series has been from the 2 energy level.
The expression for the wavelength transition can be given by:
[tex]\rm \dfrac{1}{\lambda}\;=\;R\;[\dfrac{1}{2^2}\;-\;\dfrac{1}{n^2} ][/tex]
[tex]\lambda[/tex] = Wavelength = 410 nm = 410 [tex]\rm \times\;10^-^7[/tex] cm
R = Rydbreg constant = 109678/cm
[tex]\rm \dfrac{1}{410\;\times\;10^-^7}\;=\;109678\;[\dfrac{1}{2^2}\;-\;\dfrac{1}{n^2} ][/tex]
[tex]\rm \dfrac{1}{410\;\times\;10^-^7\;\times\;109678}\;=\;\;[\dfrac{1}{2^2}\;-\;\dfrac{1}{n^2} ][/tex]
[tex]\rm 0.222\;=\;\;[\dfrac{1}{2^2}\;-\;\dfrac{1}{n^2} ][/tex]
[tex]\rm 0.222\;=\;\;[\dfrac{1}{4}\;-\;\dfrac{1}{n^2} ][/tex]
[tex]\rm 0.222\;-\;\dfrac{1}{4}\;=\;\dfrac{1}{n^2}[/tex]
[tex]\rm n^2\;=\;36.2[/tex]
n = 6
The value of n for the Hydrogen transition in the Balmer series with the wavelength 410 nm has been 6.
For information about the Balmer series, refer to the link:
https://brainly.com/question/5295294
