Problem 59
Question
The energies of photons emitted by one-electron atoms and ions fit the equation $$E=\left(2.18 \times 10^{-18} \mathrm{J}\right) \mathrm{Z}^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$$ where \(Z\) is the atomic number, \(n_{1}\) and \(n_{2}\) are positive integers, and \(n_{2}>n_{1} .\) Is the cmission associated with the \(n=2 \rightarrow n=1\) transition in a one-electron ion ever in the visible region? Why or why not?
Step-by-Step Solution
Verified Answer
Answer:
No, the emission associated with an n=2 to n=1 transition in one-electron ions does not correspond to the visible region of the spectrum. The calculation shows that the energy of the transition is higher than the maximum energy of the visible spectrum.
1Step 1: Identify the energy range of the visible region
The visible region of the electromagnetic spectrum corresponds to wavelengths between 400 nm and 700 nm. To compare the transition energy with this range, we will convert the wavelength range to an energy range using the relation, \(E=\frac{hc}{\lambda}\), where \(h\) is Planck's constant (\(6.63\times10^{-34}\) J s) and \(c\) is the speed of light (\(3\times10^8\) m/s).
$$
E_{\min} = \frac{hc}{700 \times 10^{-9}\text{m}} \quad\text{and}\quad E_{\max} = \frac{hc}{400 \times 10^{-9}\text{m}}
$$
2Step 2: Calculate the minimum energy for the n=2 to n=1 transition
For one-electron ions, the minimum energy for the \(n=2 \rightarrow n=1\) transition occurs when \(Z=1\) (i.e., in a hydrogen atom). Plug these values into the provided equation for photon energy:
$$
E_{\text{trans}} = \left( 2.18\times 10^{-18}\;\text{J}\right)(1)^2\left( \frac{1}{1^2} - \frac{1}{2^2} \right) \\
E_{\text{trans}} = 2.18\times 10^{-18}\;\text{J}\cdot\left(1 - \frac{1}{4}\right)
$$
3Step 3: Calculate the energy difference
Subtract the minimum transition energy from the maximum energy of the visible spectrum to determine if there is any overlap between the two.
$$
\Delta E = E_{\max} - E_{\text{trans}}
$$Compute this value.
4Step 4: Determine if the transition falls in the visible range
If \(\Delta E\) obtained in step 3 is positive and the minimum energy for the \(n=2 \rightarrow n=1\) transition is within the energy range of the visible spectrum, then we can conclude that the transition energy is in the visible region. Otherwise, the transition does not fall within the visible range.
Compare the calculated value of \(E_{\text{trans}}\) with \(E_{\min}\) and \(E_{\max}\) to determine whether this condition is met or not.
Key Concepts
One-Electron AtomsEnergy TransitionVisible SpectrumAtomic Number
One-Electron Atoms
One-electron atoms are systems where only a single electron orbits the nucleus. The most common example is the hydrogen atom. However, there are also ions such as \({\text{He}}^+,\ {\text{Li}}^{2+}\), or \({\text{Be}}^{3+} \) where additional electrons are removed, leaving just one electron behind.
The simplicity of one-electron atoms makes them perfect for studying atomic physics and quantum mechanics. Since they have only one electron, the forces involved are less complicated compared to multi-electron atoms.
The simplicity of one-electron atoms makes them perfect for studying atomic physics and quantum mechanics. Since they have only one electron, the forces involved are less complicated compared to multi-electron atoms.
- This simplicity allows physicists to derive exact solutions for the electron's behavior and energy levels.
- The energy levels in these systems are described by discrete values. These quantized energy levels arise from the electron's restricted orbits around the nucleus.
- Transitions between energy levels lead to either absorption or emission of photons.
Energy Transition
Energy transition in one-electron atoms involves an electron changing its energy level. This transition can involve either moving to a higher energy level, consuming a photon (absorption), or dropping to a lower energy level, releasing a photon (emission).
The energy associated with these transitions can be calculated using the formula: \[E = \left(2.18 \times 10^{-18} \mathrm{J}\right) \mathrm{Z}^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\]Where:
Understanding how and why these transitions occur provides insight into the way atoms interact with light and energy.
The energy associated with these transitions can be calculated using the formula: \[E = \left(2.18 \times 10^{-18} \mathrm{J}\right) \mathrm{Z}^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\]Where:
- \(Z\) is the atomic number, representing the number of protons in the nucleus.
- \(n_1\) and \(n_2\) are energy levels with \(n_2 > n_1\).
Understanding how and why these transitions occur provides insight into the way atoms interact with light and energy.
Visible Spectrum
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. It ranges approximately from 400 nm to 700 nm in wavelength.
When discussing photon emissions, the visible spectrum is essential because it tells us if a transition can be seen as visible light. The energy of a photon in this range is determined by converting the wavelength to energy using the equation:\[E=\frac{hc}{\lambda}\]Where,
While many energy transitions result in wavelengths outside the visible spectrum, knowing when they fall within this range helps identify whether they can be further physically observed.
When discussing photon emissions, the visible spectrum is essential because it tells us if a transition can be seen as visible light. The energy of a photon in this range is determined by converting the wavelength to energy using the equation:\[E=\frac{hc}{\lambda}\]Where,
- \(h = 6.63 \times 10^{-34}\) Js is Planck's constant,
- \(c = 3 \times 10^8\) m/s is the speed of light,
- and \(\lambda\) is the wavelength.
While many energy transitions result in wavelengths outside the visible spectrum, knowing when they fall within this range helps identify whether they can be further physically observed.
Atomic Number
The atomic number \(Z\) is a fundamental property of atoms representing the number of protons found in the nucleus. It provides crucial information about the identity of an element on the periodic table.
Atomic number serves as the cornerstone in the discussion of atomic properties and behaviors, making it crucial for both theoretical and experimental chemistry and physics.
- A higher atomic number generally means a more positively charged nucleus, which can influence how electrons orbit the nucleus.
- In the formula for photon energy in one-electron atoms, \(Z\) directly affects the energy, since it appears as \(Z^2\).
- This means elements with a higher atomic number (and therefore, more protons) will usually have higher energy transitions compared to those with lower \(Z\).
Atomic number serves as the cornerstone in the discussion of atomic properties and behaviors, making it crucial for both theoretical and experimental chemistry and physics.
Other exercises in this chapter
Problem 57
What is the wavelength of the photons emitted by hydrogen atoms when they undergo \(n=4 \rightarrow n=3\) transitions? In which region of the electromagnetic sp
View solution Problem 58
What is the frequency of the photons emitted by hydrogen atoms when they undergo \(n=5 \rightarrow n=3\) transitions? In which region of the electromagnetic spe
View solution Problem 61
By absorbing different wavelengths of light, an electron in a hydrogen atom undergoes a transition from \(n=2\) to \(n=3\) and then from \(n=3\) to \(n=4\) a. A
View solution Problem 62
The hydrogen atomic emission spectrum includes a UV line with a wavelength of \(92.3 \mathrm{nm}\) a. Is this line associated with a transition between differen
View solution