Answer:
[tex]1.936\times 10^{-18}\ \text{J}[/tex]
Explanation:
[tex]R_h[/tex] = Rydberg constant = [tex]2.178\times 10^{-18}\ \text{J}[/tex]
[tex]n_i[/tex] = Initial shell = 3
[tex]n_f[/tex] = Final shell = 1
We have the relation
[tex]\Delta E=R_h(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})\\\Rightarrow \Delta E=2.178\times 10^{-18}(\dfrac{1}{1^2}-\dfrac{1}{3^2})\\\Rightarrow \Delta E=1.936\times 10^{-18}\ \text{J}[/tex]
The energy of the photon emitted here is [tex]1.936\times 10^{-18}\ \text{J}[/tex].
