An electron in a hydrogen atom relaxes to the n = 4 level, emitti...

Question

An electron in a hydrogen atom relaxes to the n = 4 level, emitting
light of 114 THz. What is the value of n for the level in
which the electron originated?

Accepted Answer

Welcome back everyone in this example, we're told that an electron relaxes to the fourth energy level, releasing a photon with the frequency of 69 Terror hurts. And we need to determine the energy level that the electron originates from. So what that means is that we need to find the initial energy level. However, we do know that based on the prompt, because we recall that end here tells us our energy level or our shell number is at the final energy level being four. So we would say N. F. Is equal to four. And what we should recognize is that from going to one energy level to the next. Our electron goes through a transfer of energy. And in this transfer of energy because our electron prefers to be at lower energy levels, This transfer of energy is expressed through a photon. And so what we need to recall is how to find the total energy of this photon. And so we would recall that our energy of our photon can be calculated by taking Plank's constant, which we recall is represented by age multiplied by our frequency. And we'll actually use a different color for frequency. So we'll say times frequency here. And according to the prompt, we have a frequency equal to 69 terror hurts. But we want to recall that frequency can be expressed in units of Hertz, which we should recognize. Our equivalent to inverse seconds. Now plugging in our unit for planks constant. We would recall that it's equal to a value of 6.6-6 times. And we'll put an X. So times 10 To the negative 34th power. And we have units of jewels times seconds. So because we have units of seconds here for planks constant, we want to cancel out seconds so that we're left with jewels as our final unit for the energy of our photon, meaning we want our frequency to be in units of inverse seconds. And so when plugging in our frequency, we are going to recall that our prefix tera tells us that we have 69 times 10 To the 12th power inverse seconds. And so that would be how we convert to inverse seconds for our frequency unit. And this will allow us to cancel out our unit of seconds in the numerator with inverse seconds here in the denominator. And so what we would get for the energy of our photon is a value equal to 4.3 or 4.5. Sorry about that 719 times 10 to the end. We use the multiplication symbol here, So times 10 to the negative 20th power jewels. So now we have the energy of our photon that emits from our electron as it transitions in energy. And we should recognize that in the prompt, it says that this photon is released. And so we have a release of energy which we should recall is going to be described as a negative. So we'll say negative transfer of energy. And so what this means is that for the energy of our photon calculation that we have here, we can recall that this is going to equal the change in energy of our atom which is releasing energy. And so this would be a negative value as far as our energy of our photon. And so we would say that we have a value equal to negative, 4.5719 times 10 to the negative times To the negative 20th power jewels. And now we want to recall our formula for calculating the change in energy of our atom. And so we should recall that the change in energy of our atom can be calculated by taking negative one times the electron energy for a hydrogen atom which we should recall is represented by the symbol R H sub H, which is then multiplied by one over the final energy level which is squared subtracted from one over the initial energy level squared and it must be final minus initial. Here. Now as we stated, recall that R. C. Stands for the electron energy in a hydrogen atom and we should recognize that this has a value equal to 2.18 times 10 to the negative 18th power and units of jewels. So we want to plug in what we know for this formula to find our initial energy level that our electron originates from. So we know that the change in energy of our atom is going to be negative 4.5719 times 10 to the negative 20th power jules. This is set equal to negative one which is multiplied by the electron energy for hydrogen atom which would be 2. times 10 to the negative 18th power jewels. Now this is for a hydrogen atom because hydrogen atoms emit photons as light, which is what our prompt is describing here. And so now we have this plugged in. We're going to multiply this by our parentheses where we have one over our final energy level squared, which according to the prompt, our final energy level is the fourth energy level. So we would say four squared And this is subtracted from one over the initial energy level squared, which is what we want to solve for. So we need to simplify this by isolating for an I squared. And to begin, we're going to divide both sides by negative 2.18 times 10 To the negative 18th power jewels. So this allows us to cancel this out on the right hand side. And we would be able to simplify to now 0.20. And on the right hand side we would just have our math for 1/4 squared. Which in our calculators will give us a value of 0.625 and then we still have minus one over the end sub I squared value. So now to further isolate foreign sub I squared. We want to go ahead and subtract 0.0625 from both sides of our equation. And so this would cancel this out on the right hand side where we can continue to simplify so that we have one over the end sub I squared value on the right or on the left hand side now. And this is going to be set equal to a value of positive 0.0415 because we would get a negative value from this. But when we multiplied by the negative one on the right hand side, we get a positive value now. And so now we have a diagonal here. And we should recall that when we have diagonals in math or in algebra we can just switch places with those diagonals. And so what that means is that we would now have our initial energy level squared equal to 1/ 0.415. And our last step is canceling out that square term. So we would take the square root of both sides to cancel out the square term where we would get that Our initial energy level is equal to a value of about 4.9. Which rounding to one sig fig we can say is a value of five. And so we would say that therefore our electron transitioned from and equals five to n equals four as the final energy level. So we went from a high energy level to a low energy level, and we should recall that this means that we have an electron emission. So for our final answer to complete the prompt, we would say that our initial energy level is at the fifth energy level, so our electron originates from the fifth energy level. So what's highlighted in yellow is our final answer. I hope that everything I reviewed was clear. But if you have any questions, please leave them down below and I will see everyone in the next practice video.

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