Chapter 38: Q. 24 (page 1114)

In the following Figure is an energy-level diagram for a simple atom. What wavelengths, in nm, appear in the atom’s (a) emission spectrum and (b) absorption spectrum?

Short Answer

Expert verified

(a) The emission spectrum wavelengths are $λ_{21} = 828 .26 nm, λ_{31} = 310 .6 nm, λ_{32} = 497 .36 nm$

(b) The absorption spectrum wavelengths are $λ_{12} = 828 .26 nm, λ_{13} = 310 .6 nm$

Step by step solution

Part (a) Step 1: Given information

The given wavelength from the transition is $λ = \frac{hc}{ΔE}$

Part (a) Step 2: Finding the wavelengths of emission spectrum

The solution can begin with a wavelength equation from the transition.

$λ = \frac{hc}{ΔE} λ_{21} = \frac{hc}{E_{2} - E_{1}} (transition 2 \to 1) λ_{21} = \frac{6 .63 \cdot 10^{- 34} \cdot 3 \cdot 10^{8}}{1 .5 \cdot 1 .6 \cdot 10^{- 19} - 0} (substitute) λ_{21} = 828 .26 nm λ_{31} = \frac{hc}{E_{3} - E_{1}} (transition 3 \to 1) λ_{31} = \frac{6 .63 \cdot 10^{- 34} \cdot 3 \cdot 10^{8}}{4 \cdot 1 .6 \cdot 10^{- 19} - 0} (substitute) λ_{31} = 310 .6 nm λ_{32} = \frac{hc}{E_{3} - E_{2}} (transition 3 \to 2) λ_{32} = \frac{6 .63 \cdot 10^{- 34} \cdot 3 \cdot 10^{8}}{4 \cdot 1 .6 \cdot 10^{- 19} - 1 .5 \cdot 1 .6 \cdot 10^{- 19}} (substitute) λ_{32} = 497 .36 nm$

Part (b) Step 1: Given information

The chemical make-up of the gas in the stellar atmosphere determines which wavelengths are absorbed.

Part (b) Step 2: Finding the wavelengths of Absorption spectrum

We only have absorption from $1 \to 2$ and $1 \to 3$ because the atom is in the n=1 state. Because the expressions are the same, the wavelengths are the same.

$1 \to 2 : λ_{12} = 828 .26 nm 1 \to 3 : λ_{13} = 310.6 nm$