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Starting with equation (10-4), show that if Δ²Ô is-1 as a photon is emitted by a diatomic molecule in a transition among rotation-vibration states, but Δℓcan be±1 . Then the allowed photon energies obey equation (10-6).

Short Answer

Expert verified

The photon’s energy is equal toEphoton=hκμ±1h2μ²¹2 .

Step by step solution

01

Calculate the energy change for a molecule

Let us calculate the energy change for a molecule changing from state n,l to state n-1, l+1, this is the photons energy.

Enl−En−l,l+1=n+12hκμ+h2×l×(l+1)2μ²¹2−n−1+12hκμ+h2×(l+1)×(l+2)2μ²¹2=hκμ+h2×(l−(1+2))×(l+1)2μ²¹2=hκμ−h2×(l+1)μ²¹2

Then let us calculate the energy change for a molecule changing from state n,l to state n-1,l-1. This is the photon’s energy.

Enl−En−l,l−1=n+12hκμ+h2×l×(l+1)2μ²¹2−n−1+12hκμ+h2×(l−1)×(l)2μ²¹2=hκμ+h2×(l+1−(l−1))×(1)2μ²¹2=hκμ−h2×lμ²¹2

02

Conclusion.

Thus the photon’s energy is equal to

Ephoton=hκμ±1h2μ²¹2I=1, 2, 3,..........

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