Question

A photon of 300 nm absorbed by a gas and re-emited as two photons one having wavelength 760 nm what would be the wavelength of second photon

Answer 1

Use conservation of energy

Energy of incident (absorbed) photon = hc/λ

12400 eV- A° / 3000 = 4.1 eV

Energy of one photon = 12400 / 7600 = 1.6 eV

So energy of second photon will be = 4.1 - 1.6= 2.5 eV

So wavelength = 12400/ 2.5 = 4960 A°

Answer 2

496 nm is the final answer.

Explanation:

• Since energy values are additives.
• E = E₁ + E₂.
• ​hc/ ​λ = hc/ ​λ₁ +hc/ ​λ₂
• ⇒1/300 = 1/760 + 1/​λ₂
• ⇒1/​λ₂ = 1/300 - 1/760
• ⇒λ=495.6 nm=496 nm​.

• Due to the law of conservation of energy, we recognize that the strength of the photon absorbed and the energies of the photons emitted are going to be the same.

We will adjust the equation for the De Broglie wavelength to healthy our wishes and use it along with the law of conservation of energies.

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