Atomic StructureHard

Question

A dye emits 50% of the absorbed energy as fluorescence. If the number of quanta absorbed and emitted out is in the ratio 1: 2 and it absorbs the radiation of wavelength ‘x’ Å, then the wavelength of the emitted radiation will be

Options

A.x Å
B.0.5x Å
C.4x Å
D.0.25x Å

Solution

$E_{\text{emitted}} = \frac{50}{100} \times E_{\text{absorbed}}$

Or, $n_{1}.\frac{hc}{\lambda_{1}} = \frac{1}{2} \times n_{2} \times \frac{hc}{\lambda_{2}}$

Or, $\frac{\lambda_{1}}{\lambda_{2}} = \frac{2}{1} \times \frac{n_{1}}{n_{2}} = \frac{2}{1} \times \frac{2}{1} \Rightarrow \frac{\lambda_{1}}{x\overset{o}{A}} = \frac{4}{1} \Rightarrow \lambda_{1} = 4x\overset{o}{A}$

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