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Standard XII

Physics

Intensity Explanation in YDSE

A certain dye...

Question

A certain dye absorbs light of $λ = 4530 A^{0}$ and then fluoroscence light of $5080 A^{0}$ .Assuming that under given conditions 47% of the absorbed energy is reemited out as fluoroscence the ratio of quanta emitted out to the number of quanta absorbed.

0.527

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Solution

The correct option is A
0.527

$\frac{m_{2}}{m_{1}} = \frac{E_{1} λ_{1}}{E_{2} λ_{2}}$

$\Rightarrow \frac{5080 \times 0.97}{4580}$

$0.52$

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Similar questions

Q. A certain dye absorbs light of

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. Assuming that under given condition 47% of the absorbed energy is remitted out as fluorescence the ratio of quanta emitted out to the number of quanta absorbed is:

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A certain dye absorbs light of λ = 4530 A0 and then fluoroscence light of 5080 A0.Assuming that under given conditions 47% of the absorbed energy is reemited out as fluoroscence the ratio of quanta emitted out to the number of quanta absorbed.

The correct option is A 0.527 m2m1 = E1λ1E2λ2 ⇒ 5080 × 0.974580 0.52

A certain dye absorbs light of $λ = 4530 A^{0}$ and then fluoroscence light of $5080 A^{0}$ .Assuming that under given conditions 47% of the absorbed energy is reemited out as fluoroscence the ratio of quanta emitted out to the number of quanta absorbed.

The correct option is A
0.527

$\frac{m_{2}}{m_{1}} = \frac{E_{1} λ_{1}}{E_{2} λ_{2}}$

$\Rightarrow \frac{5080 \times 0.97}{4580}$

$0.52$