Question

A certain dye absorbs light of $\lambda$ = 4500 $A^{\circ }$ and then fluoresces light of $\lambda$ = 5000 $A^{\circ }$. Assuming that under given conditions, 45% of the absorbed energy is re-emitted out as fluorescence, calculate the ratio emitted out to the number of quanta absorbed.

Answer 1

Wavelength of absorbed radiation $\left(\lambda \right)=4500\stackrel{0}{A}=4500\times {10}^{-10}m$

energy of photon corresponding to this radiation$=\frac{hc}{\lambda }$

$E_a==\frac{6.626\times {10}^{-34}J-S\times 3\times {10}^8\frac{m}{s}}{4500\times {10}^{-10}m}$

$E_a=4.417\times {10}^{-19}J$

Wavelength of emitted radiation $\left(\lambda \right)=5000\stackrel{0}{A}$

$E_c=\frac{hc}{\lambda }=3.976\times {10}^{-19}J$

45% of absorbed energy is being emited as flurosence

Thus,corresponding to photon absorbed energy being reemitted $=\frac{45}{100}\times 4.417\times {10}^{-19}J$

$=1.988\times {10}^{-19}J$

No of photons emitted corresponding to this energy$=\frac{energybeingemitted}{energyofphoton}$

$=\frac{1.988\times {10}^{-19}J}{3.9756\times {10}^{-19}}$

=$0.4999$

$\cong 0.5$

Ratio of quanta emitted :quanta absorbed =$0.5:1$

$=1:2$