Start typing, then use the up and down arrows to select an option from the list.

Table of contents

here in states. What is the energy of a photon in jewels released during a transition from n equals four to n equals one state in the hydrogen atom. All right, so here, what we're gonna have is we're gonna say Delta E equals negative R E times one over N squared final, minus one over and squared Initial. We'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 our when we're dealing with jewels. It's negative. 2.178 times 10 to the negative 18 jewels. Here we start off at N equals four. We're going down to n equals one. So our final and value is one so that be one squared and our initial would be 1/4 squared. So this comes out to be negative. 2.178 times 10 to the negative. 18 jewels. When we do 1/1 squared, minus 1/4 squared, that comes out 2.9375 When these two numbers multiply with each other, that's going to give us the energy involved with this electron transitioning transitioning from an n equals four to In n equals one state. So that comes out to be negative. 2.0 four 19 times 10 to the negative 18 jewels here, let's just do it in terms of 366 So negative 2.4 times 10 to the negative 18 jewels. In this question, all we have are shell numbers. But there are only one Sig Figs. So here, at the more comfortable, giving it three sig figs. Right, So here, negative 2.4 times 10 to the negative. 18 jewels are released as the electron moves from the fourth orbital level to the first orbital level.

Related Videos

Related Practice

08:10

08:12

06:08

02:59

03:37

04:50

02:08

© 1996–2023 Pearson All rights reserved.