Question

Certain elements emit light of a specific wavelength when they are burned. Historically, chemists used such emission wavelengths to determine whether specific elements were present in a sample. Characteristic wavelengths for some of the elements are given in the following table: \(\begin{array}{llll}\mathrm{Ag} & 328.1 \mathrm{nm} & \mathrm{Fe} & 372.0 \mathrm{nm} \\ \mathrm{Au} & 267.6 \mathrm{nm} & \mathrm{K} & 404.7 \mathrm{nm} \\ \mathrm{Ba} & 455.4 \mathrm{nm} & \mathrm{Mg} & 285.2 \mathrm{nm} \\ \mathrm{Ca} & 422.7 \mathrm{nm} & \mathrm{Na} & 589.6 \mathrm{nm} \\ \mathrm{Cu} & 324.8 \mathrm{nm} & \mathrm{Ni} & 341.5 \mathrm{nm}\end{array}\) (a) Determine which elements emit radiation in the visible part of the spectrum. (b) Which element emits photons of highest energy? Of lowest energy? (c) When burned, a sample of an unknown substance is found to emit light of frequency \(6.59 \times 10^{14} \mathrm{~s}^{-1} .\) Which of these elements is probably in the sample?

Short answer

(a) Elements that emit radiation in the visible part of the spectrum are Ag, Fe, K, Ba, Ca, Na, and Ni. (b) The element with the highest energy emission is Au, and the element with the lowest energy emission is Na. (c) The unknown substance most likely contains the element Barium (Ba).

Step-by-step solution

Recall the relation between wavelength, frequency, and speed of light

The wavelength (λ) and frequency (ν) of electromagnetic waves are related to the speed of light (c) as follows: \[c = λν\] where c = \(3.00 \times 10^8 \mathrm{ m/s}\). As the given wavelengths are in nm, we will convert them to meters to work with SI units: \[1 nm = 1\times10^{-9} m\]

Calculate frequencies of the emission

Since we will be comparing frequencies, we can calculate the frequencies of the elements using the given wavelengths and the equation \(c = λν\). To find the frequency (ν), we can rearrange the equation to get: \[ν = \frac{c}{λ}\] For each element, we will calculate its frequency (in s^{-1}): 1. Ag: \(ν_{Ag} = \frac{3.00 \times 10^8 m/s}{328.1\times10^{-9} m} ≈ 9.14\times10^{14}\mathrm{ s^{-1}}\) 2. Fe: \(ν_{Fe} = \frac{3.00 \times 10^8 m/s}{372.0\times10^{-9} m} ≈ 8.06\times10^{14}\mathrm{ s^{-1}}\) 3. Au: \(ν_{Au} = \frac{3.00 \times 10^8 m/s}{267.6\times10^{-9} m} ≈ 1.12\times10^{15}\mathrm{ s^{-1}}\) 4. K: \(ν_{K} = \frac{3.00 \times 10^8 m/s}{404.7\times10^{-9} m} ≈ 7.41\times10^{14}\mathrm{ s^{-1}}\) 5. Ba: \(ν_{Ba} = \frac{3.00 \times 10^8 m/s}{455.4\times10^{-9} m} ≈ 6.59\times10^{14}\mathrm{ s^{-1}}\) 6. Mg: \(ν_{Mg} = \frac{3.00 \times 10^8 m/s}{285.2\times10^{-9} m} ≈ 1.05\times10^{15}\mathrm{ s^{-1}}\) 7. Ca: \(ν_{Ca} = \frac{3.00 \times 10^8 m/s}{422.7\times10^{-9} m} ≈ 7.10\times10^{14}\mathrm{ s^{-1}}\) 8. Na: \(ν_{Na} = \frac{3.00 \times 10^8 m/s}{589.6\times10^{-9} m} ≈ 5.09\times10^{14}\mathrm{ s^{-1}}\) 9. Cu: \(ν_{Cu} = \frac{3.00 \times 10^8 m/s}{324.8\times10^{-9} m} ≈ 9.24\times10^{14}\mathrm{ s^{-1}}\) 10. Ni: \(ν_{Ni} = \frac{3.00 \times 10^8 m/s}{341.5\times10^{-9} m} ≈ 8.79\times10^{14}\mathrm{ s^{-1}}\)

Determine visibility of the radiation

The visible spectrum has wavelengths ranging from approximately 400 to 700 nm. From the wavelength table above, one can directly see that many of the elements have wavelengths within the visible spectrum. Specifically, the elements with wavelengths in the visible range are: - Ag - Fe - K - Ba - Ca - Na - Ni

Find the highest and lowest energy emissions

The energy of emitted photons is inversely proportional to their wavelength, with higher frequency having higher energy. So, the element with the lowest wavelength has the highest energy emission, and the element with the highest wavelength has the lowest energy emission. From the table above: - Highest energy: Au (267.6 nm) - Lowest energy: Na (589.6 nm)

Determine the unknown substance

We are given that the unknown substance has an emission frequency of \(6.59 \times 10^{14} \mathrm{ s^{-1}}\). We will now compare this value with the frequency of each element we calculated in the second step to determine the closest match. Comparing the frequencies, the closest match to the given frequency is Ba with a frequency of \(6.59\times10^{14}\mathrm{ s^{-1}}\). So, the unknown substance most likely contains the element Barium (Ba). In summary: (a) Elements that emit radiation in the visible part of the spectrum: - Ag - Fe - K - Ba - Ca - Na - Ni (b) Element with the highest energy emission: Au, with lowest energy emission: Na (c) The unknown substance most likely contains the element Barium (Ba).

Key concepts

Wavelength and Frequency Relationship

Understanding the relationship between wavelength and frequency is foundational to grasping how light behaves. This connection is explained by a simple, yet fundamental equation in physics: \( c = \lambdau \). This equation states that the speed of light (\( c \)), which is around \(3.00 \times 10^8 \) meters per second (m/s), is equal to the product of the light wave's wavelength (\( \lambda \)) and its frequency (\( u \)).

These two properties of a wave are inversely related. As the wavelength increases, the frequency decreases and vice versa. For instance, if we have a wave with a wavelength of 500 nanometers (nm), we can calculate its frequency by rearranging the formula: \( u = \frac{c}{\lambda} \). By substituting the speed of light and the wavelength into this equation, we obtain the frequency of the wave, which will give us an idea of its energy and where it might fall on the electromagnetic spectrum. This foundational concept is crucial for analyzing light emitted by different elements and for a wide array of applications in science and technology.

By applying this relationship, we can determine, for example, that the light emitted by Barium (Ba) with a wavelength of 455.4 nm has a corresponding frequency calculated by the equation, giving us insight into the type of radiation Ba emits when it is burned.

Visible Spectrum Radiation

The visible spectrum is the portion of the electromagnetic spectrum that the human eye can perceive. This range spans wavelengths from approximately 400 to 700 nm, comprising the colors violet, indigo, blue, green, yellow, orange, and red. Each color has its own range within the visible spectrum, with violet having the shortest wavelength (around 400 nm) and red having the longest (around 700 nm).

The significance of this spectrum is evident when studying the emission spectra of elements. When an element burns and emits light in the visible range, we can see it with the naked eye. For instance, from our table, elements such as Silver (Ag), Iron (Fe), and Sodium (Na), amongst others, have characteristic wavelengths that fall within the visible spectrum. This means that their flame tests would reveal distinctive colors unique to each element, which historically served as an essential method for chemical identification and analysis. Moreover, understanding what part of the visible spectrum these emissions lie in can help in applications such as spectroscopy, where analyzing light can reveal information about the composition of celestial bodies or chemical substances.

Identifying the color of light emitted by an element can be a practical and visually engaging aspect of learning about the electromagnetic spectrum and the nature of light.

Photon Energy Calculation

The energy of a photon is directly related to the frequency of the light wave, and this relationship is described by the equation \( E = hu \), where \( E \) is the energy of the photon, \( h \) is Planck's constant (approximately \(6.626 \times 10^{-34}\) Joule seconds), and \( u \) is the frequency of the light. Higher frequency (thus lower wavelength) photons have more energy than lower frequency (higher wavelength) photons.

Using this formula, we can calculate energy differences between light emitted by different elements. For example, with the wavelength information given for Gold (Au) at 267.6 nm, we can calculate a high frequency and thus determine that Au photons have high energy. Conversely, Sodium (Na) photons, with a wavelength of 589.6 nm, would have lower energy because they are at the other end of the energy spectrum.

The concept of photon energy is not only important in theoretical physics but plays a significant role in understanding chemical reactions, electronic devices such as LEDs and lasers, and even the mechanisms of photosynthesis in plants. By understanding how to calculate photon energy, we gain deeper insight into the interaction of light with matter and can apply this knowledge across various scientific and technological fields.