Ncert Exercises & Solutions

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

Frequency of radiation (ν),
$v = \frac{1}{2.0 x 10^{- 9} s}$
ν = 5.0 × $10^{8} s^{- 1}$
Energy (E) of source = Nhν
Where,
N = number of photons emitted
h = Planck’s constant ν =
frequency of radiation
Substituting the values in the given expression of (E):
E = (2.5 × $10^{15}$ ) (6.626 × $10^{- 34}$ Js) (5.0 × $10^{8} s^{- 1}$ )

E = 8.282 × $10^{- 10}$ J
Hence, the energy of the source (E) is 8.282 × $10^{- 10}$ J.

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