Answer to Question #17913 in Physical Chemistry for Aidan242
E(n) = -Z2*R/n2, where Z = 2;
Energy of an absorbed photon between the levels n=1 and m=4
E(m,n) = Z2*R*(1/n2 - 1/m2)
We will take R = 10973731.6 m-1, Rydberg constant for H, in reciprocal meters.
lambda = h*c/E = 1/Z2*R*(1/n2 - 1/m2)) = 1/(22*10,973,731.6 m-1*(1/12 - 1/42)) = 3.746*10-8 m
Answer: 3.746*10-8 meters
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