Calculate energy of transition from n=3 to n=1. What region of electromagnetic spectrum is this radiation found?

Question

15702 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-29T07:29:15

Since you did not specify, I'm going to assume you mean, hydrogen with a single electron.

$E_{ph} = 1.89$ $E_"ph"=1.89$ $eV$ $"eV"$

$λ = 655.2$ $lambda=655.2$ , this is in the optical (it's a red-pink)

The formula

$E_{ph} = h c R ({(\frac{1}{n_{lower}})}^{2} - {(\frac{1}{n_{upper}})}^{2})$ $E_"ph"=hcR((1/n_"lower")^2 - (1/n_"upper")^2)$

$h c R = 13.6$ $hcR=13.6$ $eV$ $"eV"$

$E_{ph} = 13.6$ $E_"ph"=13.6$ $eV ({(\frac{1}{2})}^{2} - {(\frac{1}{3})}^{2})$ $"eV" ((1/2)^2 -(1/3)^2)$

$E_{ph} = 13.6$ $E_"ph"=13.6$ $eV (\frac{1}{4} - \frac{1}{9})$ $"eV" (1/4 - 1/9)$

$E_{ph} = 13.6$ $E_"ph"=13.6$ $eV (\frac{5}{36}) = 1.89$ $"eV" ({5}/36)=1.89$ $eV$ $"eV"$

$E_{ph} = h (\frac{c}{λ}) = h c R \frac{5}{36}$ $E_"ph"=h(c/lambda)=hcR {5}/36$

$\frac{1}{λ} = R \frac{5}{36}$ $1/lambda=R {5}/36$

$λ = \frac{1}{R} \frac{36}{5}$ $lambda=1/R 36/5$

$\frac{1}{R}$ $1/R$ is a constant and $91$ $91$ $nm$ $"nm"$

$λ = 655.2$ $lambda=655.2$ , this is in the optical (it's a red-pink)