Here it states what is the energy of a photon in joules released during a transition from N = 4 to N = 1 state in the hydrogen atom. All right, so here what we're going to have is we're going to say
ΔE = - R E × 1 N 2 - 1 N 2We're using this form of the equation because we're dealing with energy.
All we do now is we plug in the values that we know. First of all, the Rydberg constant, which is R. When we're dealing with Joules, it's
- 2.178 × 10 -18 JoulesHere we start off at N = 4, and we're going down to N = 1, so our final N value is 1, so that be
1 2and our initial would be
1 4 2So this comes out to be
- 2.178 × 10 -18 JoulesWhen we do
1 1 2 - 1 4 2that comes out to .9375. When these two numbers multiply with each other, that's going to give us the energy involved with this electron transitioning from an N = 4 to an N = 1 state. So that comes out to be
- 2.0419 × 10 -18 joulesHere, let's just do it in terms of three sig figs, so
- 2.04 × 10 -18 joulesin this question. All we have are shell numbers, but there are only one sig figs. So here I feel more comfortable giving it three sig figs, right? So here
- 2.04 × 10 -18 joulesare released as the electron moves from the 4th orbital level to the first orbital level.