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An atom emits a photon as it makes a transition from the n = 4 state...

An atom emits a photon as it makes a transition from the n = 4 state to the n = 3 state. The energies of these two states are –1.0 eV and –1.4 eV, respectively.

(a) What is the energy of the photon?

(b) What is its frequency?

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Answer #1

(a) The energy of photon = E​​​​​​4 - E​​​​​​3 = -1 - (-1.4) = 0.4 eV

(b) Plank's constant in eV = 4.135 × 10-15 eV/Hz

Energy of photon = Plank's constant × Frequency

Frequency = ( 0.4 ) / (4.135 × 10-15) = 0.096 × 1015 = 96 × 1012

  = 96 Terra Hz

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