Problem 127 - Chemistry The Science in Context

Question

Astronomers have detected hydrogen atoms in interstellar space in the \(n=732\) excited state. Suppose an atom in this excited state undergoes a transition from \(n=732\) to \(n=731\) a. How much energy does the atom lose as a result of this transition? b. What is the wavelength of radiation corresponding to this transition? c. What kind of telescope would astronomers need in order to detect radiation of this wavelength? (Hint: It would not be one designed to capture visible light.)

Step-by-Step Solution

Answer

#Answer# a. The energy difference between the two energy levels is found using the energy levels calculated from Bohr's formula: $$\Delta E = E_{732} - E_{731}$$ b. The wavelength of the radiation corresponding to this transition is computed using the energy-wavelength relationship: $$\lambda = \frac{h c}{\Delta E}$$ c. The type of telescope required to detect this radiation depends on the calculated wavelength, which will help us determine which part of the electromagnetic spectrum it lies in. Possible telescope types include infrared, ultraviolet, radio, or X-ray telescopes, depending on the wavelength.

Step 1: Calculate the energies of both energy levels

First, we need to find the energies of the \(n=732\) state and the \(n=731\) state. The energy of an electron in the hydrogen atom is given by the Bohr’s formula: $$E_n=-\frac{13.6 eV}{n^2}$$. We will use this formula to find the energies for both states: For \(n=732\): $$E_{732} = -\frac{13.6 eV}{(732)^2}$$ For \(n=731\): $$E_{731} = -\frac{13.6 eV}{(731)^2}$$

Step 2: Calculate the energy difference

Now find the energy difference between the two energy levels: $$\Delta E = E_{732} - E_{731}$$

Step 3: Compute the wavelength of radiation corresponding to this transition

We can use the energy-wavelength relationship, which states that $$E = h c / \lambda$$, where \(h = 6.63*10^{-34} Js\) is the Planck's constant, \(c = 3*10^8 m/s\) is the speed of light, and \(\lambda\) is the wavelength. We can rearrange this equation to solve for the wavelength of radiation emitted: $$\lambda = \frac{h c}{\Delta E}$$

Step 4: Determine the type of telescope required

Once we have the wavelength, we can determine the type of telescope needed to observe the radiation corresponding to this transition. We know that it is not a telescope designed for capturing visible light, so we will need to discuss the electromagnetic spectrum and the types of telescopes suitable for detecting radiation of the calculated wavelength.

Key Concepts

Bohr's model

Bohr's model of the atom, developed by Niels Bohr in 1913, plays a crucial role in understanding atomic transitions, particularly in the hydrogen atom. In this model, electrons are said to move in defined circular orbits around the nucleus. Each orbit corresponds to a specific energy level, which is denoted by the principal quantum number, \( n \). In the context of our exercise, the hydrogen atom's electron transitions from an orbit with \( n=732 \) to \( n=731 \). Using Bohr's formula, \( E_n = -\frac{13.6 \text{ eV}}{n^2} \), we can calculate the specific energies for each level. As the electron transitions to a lower energy level, it emits energy in the form of electromagnetic radiation. This quantum leap and the corresponding energy release are fundamental to Bohr’s model.

energy levels

Energy levels in an atom are the specific energies that an electron can have when occupying a particular orbit. These levels are quantized, meaning electrons can only exist at certain, discrete energy values, not between them. In a hydrogen atom, the energy levels can be calculated using the formula from Bohr's model. As an electron moves from a higher energy level to a lower one, it emits energy in quantized amounts, known as photons. In the problem we're examining, the electron transitions from \( n=732 \) to \( n=731 \). The energy loss is calculated by finding the difference between the two energy levels using Bohr's formula, \( \Delta E = E_{732} - E_{731} \). Calculating this helps us understand the energy change and photon emission, which directly relates to the wavelength of emitted radiation.

electromagnetic spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from gamma rays to radio waves. Each type of radiation varies in wavelength and frequency, with visible light occupying only a small part of the spectrum. When a hydrogen atom undergoes a transition between energy levels, the emitted radiation can fall anywhere on this spectrum, based on the energy change. The wavelength of the radiation is given by the formula \( \lambda = \frac{h c}{\Delta E} \), where \( \Delta E \) is the energy difference, \( h \) is Planck's constant, and \( c \) is the speed of light. For the transition from \( n=732 \) to \( n=731 \), the wavelength calculated using this equation helps determine the type of electromagnetic radiation emitted and the best detection method.

telescope types

Telescopes are designed to observe different parts of the electromagnetic spectrum by capturing specific types of radiation. Depending on the wavelength of radiation, different telescopes are employed:

• Optical telescopes: These are for visible light and include refracting and reflecting types.
• Radio telescopes: Used to detect radio waves with typically large parabolic dishes.
• Infrared telescopes: Have detectors to capture heat radiation and require cooling to reduce background noise.
• Ultraviolet telescopes: These need to be placed in orbit above the Earth to avoid atmospheric absorption.
• X-ray and Gamma-ray telescopes: Also located in space due to their high-frequency radiation.

In the given exercise, the wavelength determined from the \( n=732 \) to \( n=731 \) transition isn't suited for a traditional visible light telescope. Understanding the specific region of the spectrum the radiation falls into will aid in selecting the appropriate telescope type to observe it clearly.