Quick Question on the energy of a photon

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A free electron is captured by a stationary proton, forming an excited hydrogen atom and emitting two photons: the first photon with energy Ea and a second photon of 10.2 eV. The second photon corresponds to the transition from the n=2 state to the ground state (n=1). The energy of the first photon must be greater than 10.2 eV, as it represents the transition from an unbound state (n=infinity) to n=2. The ionization energy of hydrogen is 13.6 eV, which provides context for understanding the energy levels involved. The discussion highlights the challenge of calculating the energy of the first photon based on the transitions between energy states.
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Homework Statement



A free electron with negligible kinetic energy is captured by a stationary proton to form an excited state of the hydrogen atom. During this process a photon of energy Ea is emitted, followed shortly by another photon of energy 10.2 eV. No further photons are emitted. The ionization energy of hydrogen is 13.6 eV

What is the energy Ea of the first photon?

Homework Equations


Ei - Ef = hf

The Attempt at a Solution


I am very... troubled by this step. I know that when the second photon is emitted, of energy 10.2 eV, the photon goes from the n = 1 state (ground state) to n=2 (first excited state) ... but how do I go about finding the energy Ea of the first photon? I believe it should have greater energy than the second one according to the equation above. I don't need an answer.. just something to ponder about because I'm rather stuck. The wavelength of the 10.2 eV photon is 121.875 nm if that helps.:rolleyes:
 
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The first electron was unbound with negligible kinetic energy. It went from n=infinity to n=2, and the second went from n=2 to n=1. Why should the first emission have greater energy?
 
I was maybe thinking the energy of an electron as it goes from one state to the next... but now I understand. Thank you so much ^_^
 

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