Question

the longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each radiation and energy difference between two excited states.

Answer 1

λ1 = 589 nm = 5.89X10-9m

therefore v1=c/ λ1 = 3x108/589x10-9 = 5.093x1014/sec

λ2 = 589.6nm = 589.6x10-9m

therefore v2 = c/ λ2 = 3x108/589.6x10-9

= 5.088x1014/sec

ΔE=E2-E1 = h(v2-v1)

= (6.626x10-34)(5.093x-5.088)x1014

= 3.31x10-22J

Answer 2

λ1 = 589 nm = 5.89X10-9m

therefore v1=c/ λ1 = 3x108/589x10-9 = 5.093x1014/sec

λ2 = 589.6nm = 589.6x10-9m

therefore v2 = c/ λ2 = 3x108/589.6x10-9

= 5.088x1014/sec

ΔE=E2-E1 = h(v2-v1)

= (6.626x10-34)(5.093x-5.088)x1014

= 3.31x10-22 J

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