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What is the energy of a photon that, when absorbed by a hydrogen atom, could cause an electronic transition from (a) the $n=2$ state to the $n=5$ state and $(b)$ the $n=4$ state to the $n=6$ state?

a) 2.86 $\mathrm{eV}$

b) 0.472 $\mathrm{eV}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

to find the energy of the photons that is required to make this transition. We can use the equation that says that the energy required of the photons E is equal to negative 13.6 electron volts, which comes from the ground state energy of hydrogen multiplied by the ratio of one over the final orbital level inside death. So in our case, this would be five. This is incidents squared minus one over. Inside I squared, which, in part a were told us to plugging these values in. We find that the energy of the photon that is required is equal to 2.86 electron volts. For part B, we're going to do the exact same process. Except for this time we're going between the n equals four unequal six orbital. So once again, we have the same equation. Not a bad idea to ride it out again. Writing it down will help us remember it. Negative 13.6 CB again. That is the ground state energy of the hydrogen atom multiplied by one over inside that squared. In this case, it's six minus one over inside vice where, and you might assume just intuitively since this, since these orbital's air closer together and they're higher orbital's, that it's not gonna take as much energy. And this assumption would be correct. Plugging these values in and carrying out this operation, we find that this is equal to 0.472 electron volts you can box that in is our solution for B.

University of Kansas