Atomic StructureEasy
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, $\frac{\lambda_{1}}{\lambda_{2}} = \frac{2}{1} \times \frac{n_{1}}{n_{2}} = \frac{2}{1} \times \frac{2}{1} \Rightarrow \frac{\lambda_{1}}{x\overset{o}{A}} = \frac{4}{1} \Rightarrow \lambda_{1} = 4x\overset{o}{A}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
In different resonating form of a molecule what should be identical :-...When electron jumps from the fourth orbit to the second orbit in He+ ion, the radiation emitted out will fall in...The uncertainty in position and velocity of the particle are 0.1 nm and 5.27 × 10-27 ms-1 respectively then the mas...The difference between the wave number of 1st line of Balmer series and last line of paschen series for Li2+ ion is :...The radius of the second orbit of H-atom is equal to the radius of...