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Question

A dye absorbs a photon of wavelength lambda and re-emits the same energy into two photons of wavelengths of λ1 and λ2 respectively. The wavelength lambda is related with λ1 and λ2 as:
  1. λ=λ1+λ2λ1λ2
  2. λ=λ1λ2λ1+λ2
  3. λ=λ12λ22λ1+λ2
  4. λ=λ1λ2(λ1+λ2)2

A
λ=λ1λ2λ1+λ2
B
λ=λ1λ2(λ1+λ2)2
C
λ=λ1+λ2λ1λ2
D
λ=λ12λ22λ1+λ2
Solution
Verified by Toppr

Let E be the energy of photon with wavelength λ
E1 be the energy of photon with wavelength λ1
E2 be the energy of photon with wavelength λ2
E=E1+E2

hcλ=hcλ1+hcλ2

1λ=1λ1+1λ2

1λ=λ1+λ2λ1λ2

λ=λ1λ2λ1+λ2

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