2.2: Question 2.E.26 PASS - Bohr Model, quantized energy change

Exercise \(\PageIndex{2.E.26}\)

Using the Bohr model,

a. determine the change in energy (in joules) when the electron in a Li²⁺ ion moves from the n = 2 to the n = 1 orbit.

b. What is the energy (in joules) of the photon produced?

Answer

\[
a. \Delta \mathrm{E}=-1.47 \times 10^{-17} \mathrm{~J}
\]

\[
b. \mathrm{E_{photon}}=1.47 \times 10^{-17} \mathrm{~J}
\]

See the Bohr Model (see LibreText section 2.2)

Strategy Map

Step	Hint
1. Identify the information (data and/or variables) given in the question.	Recall the Bohr Model (see LibreText section 2.2) ni=2, nf=1 (initial and final energy levels)
2. Identify what the question is asking for; what variable would represent it in a formula.	Looking for ΔE (total energy change)
3. Choose a formula that allows you to plug in your given information. Fill in any known constant values.	\[ \Delta \mathrm{E}=-2.18 \times 10^{-18} J\left(\frac{Z^2}{n_f^2}-\frac{Z^2}{n_i^2}\right) \] to check if using the correct formula; does this formula help you? Are you closer to finding the desired variable? ni=2 nf=1 ΔE=? Rydberg's Constant (RH)=2.18x10^-17 Z=Atomic Number=3 **Note that the atomic number is found on the Periodic Table, most questions consider 'hydrogen's electron'; because hydrogen has an atomic number of Z=1, the formula is often simplified to: \[ \Delta \mathrm{E}=-R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \]
4. Calculate the final answer, do not forget about sign, units and significant figures.	Be sure to follow typical math rules such as order of operations Recall the Order of Operations (see LibreText section 1.4) Common mistakes are made when filling in the formula. When using nonhydrogen atoms, students often forget to include the atomic number (Z) in the numerator of both fractions. In this example, the atomic number is for Li^2^+ is 3.

Solution

\[
a. \Delta \mathrm{E}=-R_H\left(\frac{Z^2}{n_f^2}-\frac{Z^2}{n_i^2}\right) \\
\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{Z^2}{n_f^2}-\frac{Z^2}{n_i^2}\right) \\
\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{3^2}{1^2}-\frac{3^2}{2^2}\right) \\
\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{9}{1}-\frac{9}{4}\right) \\
\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{27}{4}\right) \\
\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}(6.25) \\
\Delta \mathrm{E}=-1.47_{15} \times 10^{-17} \mathrm{~J} \\
\Delta \mathrm{E}=-1.47 \times 10^{-17} \mathrm{~J}
\]

 

\[
b. \mathrm{E_{photon}}=1.47 \times 10^{-17} \mathrm{~J}
\]

Guided Solution

Guided Solution	Hint
This is a calculation question where you use the Rydberg Equation to calculate the energy of the transition of the electron.	Recall the Bohr Model (see LibreText Example 2.2.1)
Using the Bohr model, a. determine the change in energy (in joules) when the electron in a Li^2^+ ion moves from the n = 2 to the n = 1 orbit. b. What is the energy (in joules) of the photon produced?	change in energy = ΔE "from the" = initial energy level = ni = 2 "to the" = final energy level = nf =1 energy of the photon = Ephoton = |ΔE|
Summarize what the question is asking	we are looking for the change in energy of the Li^2^+electron when it transitions from n=2 to n=1. The energy change is exactly the same (AKA quantized) amount as what is emitted (i.e., as a photon).
Predict what you expect the answer to be based on your understanding of the related chemistry concepts	Since the Li^2^+ ion is going from a higher energy orbital (n=2) to a lower energy orbital (n=1), you can expect the answer to be a negative value. When energy is emitted is it released or lost from the system; ΔE is a negative value. When energy is absorbed, it is added to the system; ΔE is a positive value
Complete Solution: PART a. Step 1 Recall the Rydberg equation: \[ \Delta \mathrm{E}=-R_H \left(\frac{Z^2}{n_f^2}-\frac{Z^2}{n_i^2}\right) \] Step 2 Input your data provided by the question: ΔE = unknown Z = atomic number of atom (Li^2^+) = 3 nf = Final energy state = 1 ni = Initial energy state = 2 RH­ = Rydberg’s Constant (2.18x10^-17 J) \[ \Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{3^2}{1^2}-\frac{3^2}{2^2}\right)\] Step 3 Start by solving inside the brackets: \[\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{9}{1}-\frac{9}{4}\right) \\ \Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{27}{4}\right) \\ \Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{~J}(6.25) \\\] Step 4 Multiply the constant (RH) and the value inside the brackets: \[\Delta \mathrm{E}=-1.47_{15} \times 10^{-17} \mathrm{~J} \\ \Delta \mathrm{E}=-1.47 \times 10^{-17} \mathrm{~J} \\\] Part b. Step 5 Recall the relationship between the energy change for the transition of an electron to the energy of the photon that is absorbed or emitted during that change energy of the photon = Ephoton = |ΔEelectron| ΔEelectron = -1.47x10^-17 J therefore Ephoton = 1.47x10^-17 J	Re: Step 1 - the Rydberg equation allows us to calculate the overall change in energy (ΔE) of an atom, given its initial (ni) and final (nf) energy values. Re: Step 2 - Common mistakes are made here, when using nonhydrogen atoms, students often forget to include the atomic number (Z) as the numerator. Re: Step 3 - Square the numerator and denominator of both fractions. Combine the fractions by subtracting them. An optional step, to convert your fraction to a decimal before the multiplication step. Re: Step 4 - notice that there is a negative sign in front of the constant value. Re: Step 5 - the energy of a photon must exactly match the energy change resulting from the transition of the electron from one energy level (orbit) to another (orbit) A photon is a quantized packet of energy When an electron transitions from a low to a high energy level, the electron must absorb energy (ΔEelectron = positive) from a photon with the exact amount of energy required for that transition When an electron transitions from a high to a low energy level, the electron must release energy (ΔEelectron = negative). The energy lost from the electron is emitted as a photon with the exact amount of energy as was released during that transition.
Check your work! Why does this answer make chemical sense?	The calculated answer for this problem was a negative value. This number makes does make chemical sense as our final energy state is smaller than our initial energy state, (1/nfinal­ – 1/ninitial) a small number subtracted from a larger number results in a positive value inside the brackets which is then multiplied by the negative Rydberg’s constant. Because the change in energy was negative this means the energy was emitted. Recall that when atoms go from a higher energy state to a lower energy state, they emit a photon of electromagnetic energy, that amount of energy is dependent on the difference of energy in the two states. Emitted energy relates to a negative calculation, whereas absorbed energy relates to a positive calculation.

(question source adapted question 6.2.11 from 6.E: Electronic Structure and Periodic Properties (Exercises) - Chemistry LibreTexts, shared under a CC BY 4.0 license, authored, remixed, and/or curated by OpenStax, original source question 27 in Ch. 6 Exercises - Chemistry 2e | OpenStax, Access for free at https://openstax.org/books/chemistry/pages/1-introduction)