Question

Consider a transition of the electron in the hydrogen atom from \(n=4\) to \(n=9 .\) (a) Is \(\Delta E\) for this process positive or negative? (b) Determine the wavelength of light that is associated with this transition. Will the light be absorbed or emitted? (c) In which portion of the electromagnetic spectrum is the light in part (b)?

Short answer

The energy change for the transition of electron from \(n=4\) to \(n=9\) in the hydrogen atom is \(\Delta E = E_{final} - E_{initial}\). Calculating the energy levels using \(E_n = -\frac{13.6\,\text{eV}}{n^2}\), we get \(E_{initial} = -\frac{13.6\,\text{eV}}{4^2}\) and \(E_{final} = -\frac{13.6\,\text{eV}}{9^2}\). Substituting the values, we find that \(\Delta E\) is positive, which means that energy is absorbed and the electron moves to a higher energy level. The wavelength of light associated with this transition can be calculated using \(\lambda = \frac{hc}{\Delta E}\). Based on the obtained wavelength, we can identify its position in the electromagnetic spectrum. Since the energy is absorbed for this transition, the light will also be absorbed and not emitted.

Step-by-step solution

Energy Change Calculation

We can calculate the energy change using the formula: ΔE = E_final - E_initial where, E_initial and E_final are the initial and final energy levels of the electron in the hydrogen atom. For a hydrogen atom, the energy levels can be calculated using the formula: E_n = \(-\frac{13.6\,\text{eV}}{n^2}\) Calculate the energy levels for n = 4 and n = 9. E_initial (n = 4) = \(-\frac{13.6\,\text{eV}}{4^2}\) E_final (n = 9) = \(-\frac{13.6\,\text{eV}}{9^2}\) Now, compute ΔE.

Determine Sign of ΔE

Calculate ΔE using the equation: ΔE = E_final - E_initial Substitute the values of E_initial and E_final and find the value of ΔE to check if it is positive or negative.

Wavelength Calculation

We can find the wavelength of light associated with this transition using the equation: \(\Delta E = \frac{hc}{\lambda}\) Where: ΔE = energy change h = Planck's constant (6.63 × 10^-34 Js) c = Speed of light (3.00 × 10^8 m/s) λ = wavelength of light Isolate λ and compute its value: \(\lambda = \frac{hc}{\Delta E}\)

Absorbed or Emitted Light

If ΔE is negative, the electron has moved to a lower energy level, and energy has been released in the form of light, meaning the light is emitted. If ΔE is positive, the electron has moved to a higher energy level, and energy has been absorbed, meaning the light is absorbed. Determine if the light is absorbed or emitted based on the sign of ΔE.

Identify the Electromagnetic Spectrum

Using the wavelength calculated in Step 3, identify the portion of the electromagnetic spectrum it falls under. The electromagnetic spectrum is divided as follows: 1. Radio waves: λ > 1m 2. Microwaves: 1m > λ > 1mm 3. Infrared: 1mm > λ > 700 nm 4. Visible light: 700 nm > λ > 400 nm 5. Ultraviolet: 400 nm > λ > 10 nm 6. X-rays: 10 nm > λ > 0.01 nm 7. Gamma rays: λ < 0.01 nm

Key concepts

Energy Level Calculation in Hydrogen Atom

Understanding how to calculate energy levels in a hydrogen atom is fundamental in the field of quantum mechanics and is especially relevant when examining electron transitions. The energy levels of an electron in a hydrogen atom are given by the formula:
\[\begin{equation} E_n = -\frac{13.6\text{ eV}}{n^2} \end{equation}\]
This equation reveals that the energy of an electron is quantized, meaning it can only occupy certain discrete levels and is inversely proportional to the square of the principal quantum number, n. The constant 13.6 eV represents the ionization energy of the hydrogen atom, which is the amount of energy required to completely remove the electron from the atom.

• The negative sign indicates that these energy levels are bound states, meaning the electron is attached to the nucleus.
• As n increases, the absolute value of energy decreases, making the level higher and less bound to the nucleus.
• Transitions between these energy levels will either release or absorb energy, corresponding to the emission or absorption of light.

Considering the transition from n=4 to n=9, we can see these principles in action. Calculation shows that the electron is moving to a higher energy level (as n increases from 4 to 9). Therefore, such a transition requires energy absorption.

Wavelength Calculation of Emitted or Absorbed Light

The relationship between energy changes and light wavelength during electron transitions in a hydrogen atom is described by the formula:
\[\begin{equation} \( \)Delta E = \frac{hc}{\lambda} \end{equation}\]
In this equation, \( \Delta E \) represents the energy change during the transition, h is Planck's constant (approximately \( 6.63 \times 10^{-34} \) J s), c is the speed of light in a vacuum (approximately \( 3.00 \times 10^{8} \) m/s), and \( \lambda \) is the wavelength of the emitted or absorbed light.

• When \( \Delta E \) is positive, the electron is moving to a higher energy level, and light is absorbed.
• Conversely, a negative \( \Delta E \) implies a move to a lower energy level with light being emitted.
• By rearranging the equation, \( \lambda \) can be calculated from a known energy change, providing insight into the type of light involved in the transition.

To calculate the exact wavelength, one can isolate \( \lambda \) to obtain:\( \lambda = \frac{hc}{\Delta E} \). This relation allows us to convert the energy difference between levels directly into a wavelength, thereby indicating the type of electromagnetic radiation emitted or absorbed.

Electromagnetic Spectrum Classification

The electromagnetic spectrum is a continuous range of wavelengths that includes all possible frequencies of electromagnetic radiation. It is conveniently divided into several categories, each with unique characteristics:

• Radio Waves: Characterized by the longest wavelengths and lowest frequencies, used in communications.
• Microwaves: Useful in radar technology and cooking in microwave ovens.
• Infrared: Radiated by objects as heat and used in remote controls and thermal imaging.
• Visible Light: The portion of the spectrum visible to the human eye, spanning from red to violet.
• Ultraviolet: Has higher energy than visible light and can cause sunburn.
• X-rays: Penetrating radiation that can pass through objects, widely used in medical imaging.
• Gamma Rays: The highest-energy radiation, often emitted by nuclear reactions and cosmic sources.

Knowing the wavelength of light associated with electron transitions allows one to determine the place of the radiation within the electromagnetic spectrum. For example, if the energy change calculated in an electron transition corresponds to the wavelength of visible light, we know the transition involves the emission or absorption of photons that we could potentially see with our eyes.