Scientific Tutor
Chem – College: Rydberg Equation
Author: Andrew
How do we calculate the energy required to move an electron between energy levels?
The answer is the Rydberg equation. Note this Rydberg equation can only be used for hydrogen energy levels or shells. It is shown below. The equations are written two different ways. Both of the equations are the same but the second one is organized better for future demonstrated examples.
Δ E = RH((1/nI2) – (1/nf2))
| Δ E = | 1 – | 1 |
| RH | nI2 | nf2 |
The Δ E stands for the change in energy with the units of Joules (J). The RH is the Rydberg constant. It is simply the number 2.178 * 10-18 J that only applies to hydrogen and allows you to put the equation together. nI stands for the electron shell (or energy level) initial. It will be an integer like 1,2,3… The nf stands for the electron shell (or energy level) final and will also be an integer like 1,2,3…
The Rydberg equation also connects to other equations in this lesson. The most obvious one is E = h (f) in the section calculating the energy of light. Just treat Δ E and E as the same thing. The only difference between them is that Δ E can be a positive or negative number where as E has to be a positive number.
VIDEO Rydberg Equation Demonstrated Example 1: If the electron of a hydrogen atom goes from the second electron shell to the first electron shell, what is the maximum amount of energy that can be released as light? The Rydberg constant is 2.178 * 10-18 J.
Step 1:
What information are we given?
Answer:
nI = 2
nf = 1
RH = 2.178 * 10-18 J
Step 2:
What is the problem asking for?
Answer: Δ E = ?
Step 3:
What formula does the question involve?
Answer:
| Δ E = | 1 – | 1 |
| RH | nI2 | nf2 |
Step 4:
How do we fill in the numbers for the equation?
Answer:
| Δ E = | 1 – | 1 |
| 2.178 * 10-18 J | 22 | 12 |
Step 5:
How do we solve for the Δ E?
Answer: First simplify the exponents
| Δ E = | 1 – | 1 |
| 2.178 * 10-18 J | 4 | 1 |
Step 6:
Then put the fractions in decimals
| Δ E = | 0.25 – | 1 |
| 2.178 * 10-18 J |
Step 7:
Now subtract the right side
| Δ E = | – 0.75 | 1 |
| 2.178 * 10-18 J |
Step 8:
Finally multiply both sides by 2.178 * 10-18 J
| 2.178 * 10-18 J * Δ E = | – 0.75 * 2.178 * 10-18 J | 1 |
| 2.178 * 10-18 J |
Step 9:
Cross out like terms on left side
| 2.178 * 10-18 J * Δ E = | – 0.75 * 2.178 * 10-18 J | 1 |
| 2.178 * 10-18 J |
Step 10: simplify
| Δ E = | – 0.75 * 2.178 * 10-18 J | 1 |
| 1 |
Step 11: calculations
Answer: -0.75 * (2.178 * 10-18 J) = 1.6 * 10-18
| 1.6 * 10-18 = | – 0.75 * 2.178 * 10-18 J | 1 |
| 1 |
Step 12:
What is the final answer?
COMPLETE ANSWER: – 1.6 * 10-18 J
VIDEO Rydberg Equation Demonstrated Example 2: When a single hydrogen atom absorbs 2.09 * 10-18 J, and its electron starts from the first energy level, what energy level does the electron end at? The Rydberg constant is 2.178 * 10-18 J.
Step 1:
What information are we given?
Answer:
Δ E = 2.09 * 10-18 J
nI = 1
RH = 2.178 * 10-18 J
Step 2:
What is the problem asking for?
Answer: nf = ?
Step 3:
What formula does the question involve?
Answer:
| Δ E = | 1 – | 1 |
| RH | nI2 | nf2 |
Step 4:
How do we fill in the numbers for the equation?
Answer:
| 2.09 * 10-18 J = | 1 – | 1 |
| 2.178 * 10-18 J | 12 | nf2 |
Step 5:
How do we solve for nf?
Answer: First divide the left side.
| 0.96 = | 1 – | 1 |
| 12 | nf2 |
Step 6:
Then solve the exponent on the right side.
| 0.96 = | 1 – | 1 |
| 1 | nf2 |
Step 7:
Now solve the fraction on the right side.
| 0.96 = | 1 – | 1 |
| nf2 |
Step 8:
Minus 1 to both sides
| 0.96 – 1 = | 1 – 1 – | 1 |
| nf2 |
Step 9:
Cross out right side
| 0.96 – 1 = | 1 – 1 – | 1 |
| nf2 |
Step 10:
Simplify left side
| – 0.04 = | – | 1 |
| nf2 |
Step 11:
Eliminated the negatives from both sides
| 0.04 = | 1 | |
| nf2 |
Step 12:
Multiply both sides by nf2
| nf2 * 0.04 = | nf2 | 1 |
| nf2 |
Step 13:
Cross out like terms
| nf2 * 0.04 = | nf2 | 1 |
| nf2 |
Step 14:
Simplify
| nf2 * 0.04 = | 1 | |
| 1 |
Step 15:
Divide both sides by 0.04
| nf2 * 0.04 = | 1 | |
| 0.04 | 0.04 |
Step 16:
Cross out like terms
| nf2 * 0.04 = | 1 | |
| 0.04 | 0.04 |
Step 17:
Simplify
| nf2 = | 1 | |
| 0.04 |
Step 18:
Divide the right side ( 1 / 0.04 = 25 )
| nf2 = | 25 | |
| 1 |
Step 19:
Take the square root of both sides
| Sqrt nf2 = | Sqrt 25 | |
| 1 |
Step 20:
Cross out the square root and the square on the left side
| Sqrt nf2 = | Sqrt 25 | |
| 1 |
Step 21:
Simply
| nf = | Sqrt 25 | |
| 1 |
Step 22:
Take the square root of 25
| nf = | 5 | |
| 1 |
Step 23:
What is the final answer?
COMPLETE ANSWER: The 5th energy level
PRACTICE PROBLEMS: Solve the energy level (electron shell) or energy required to move electrons around in a hydrogen atom. Don’t forget to use the Rydberg constant RH of 2.178 * 10-18 J when needed.
If the electron of a hydrogen atom goes from the third electron shell to the first electron shell, what is the maximum amount of energy that can be released as light?
Answer: -1.94 * 10-18 J
If the electron of a hydrogen atom goes from the second electron shell to the fifth electron shell, what is the maximum amount of energy that can be absorbed as light?
Answer: 4.5 * 10-19 J
When a single hydrogen atom absorbs 2.04 * 10-18 J, and its electron starts from the first energy level, what energy level does the electron end at?
Answer: Fourth
When a single hydrogen atom releases -3.03 * 10-19 J, and its electron ends in the second energy level, what energy level does the electron start at?
Answer: Third
Posted in Chemistry
