Question

Calculate the value of ni for Hydrogen if a violet line from the Balmer series has a recorded wavelength of 410 nm.

Answer 1

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 $H_2$ 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

$\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$

where, R = Rydberg constant = 109678 per cm

            λ = wavelength

Here the recorded λ = 410 nm = $410 \times 14^{-7} \ cm$

∴ $\frac{1}{410 \times 10^{-7}}=109678\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$

$\frac{1}{2^2}-\frac{1}{n^2}=\frac{1}{410 \times 10^{-7}\times 109678}$

$\frac{1}{4}-\frac{1}{n^2}=0.222$

$\frac{1}{n^2}=\frac{1}{4}-0.222$

$n^2=\frac{1}{0.0276}$

$n^2=36.2$

$n^2$ ≈ 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.