Problem 109 - Chemistry The Central Science In Si Units

Question

Moseley established the concept of atomic number by studying \(X\) rays emitted by the elements. The \(X\) rays emitted by some of the elements have the following wavelengths: $$ \begin{array}{cc} \hline \text { Element } & \text { Wavelength (pm) } \\ \hline \text { Ne } & 1461 \\ \text { Ca } & 335.8 \\ \text { Zn } & 143.5 \\ \text { Zr } & 78.6 \\ \text { Sn } & 49.1 \\ \hline \end{array} $$ (a) Calculate the frequency, \(\nu,\) of the \(X\) rays emitted by each of the elements, in Hz. (b) Plot the square root of \(\nu\) versus the atomic number of the element. What do you observe about the plot? (c) Explain how the plot in part (b) allowed Moseley to predict the existence of undiscovered elements. (d) Use the result from part (b) to predict the X-ray wavelength emitted by iron. (e) A particular element emits X rays with a wavelength of \(98.0 \mathrm{pm}\). What element do you think it is?

Step-by-Step Solution

Answer

Moseley's plot shows a linear relation between \(\sqrt{\nu}\) and atomic number, predicting elements like Palladium from X-ray data.

Step 1: Calculate Frequency for Each Element

Use the formula for frequency given the wavelength of X-rays for each element: \( u = \frac{c}{\lambda} \), where \(c = 3.00 \times 10^8\) m/s is the speed of light. Convert the wavelength from picometers (pm) to meters (since 1 pm = \(1 \times 10^{-12}\) m), and then calculate the frequency \(u\) in Hz for each element:- Neon (Ne): \(u = \frac{3.00 \times 10^8}{1461 \times 10^{-12}} = 2.05 \times 10^{17}\) Hz- Calcium (Ca): \(u = \frac{3.00 \times 10^8}{335.8 \times 10^{-12}} = 8.93 \times 10^{17}\) Hz- Zinc (Zn): \(u = \frac{3.00 \times 10^8}{143.5 \times 10^{-12}} = 2.09 \times 10^{18}\) Hz- Zirconium (Zr): \(u = \frac{3.00 \times 10^8}{78.6 \times 10^{-12}} = 3.82 \times 10^{18}\) Hz- Tin (Sn): \(u = \frac{3.00 \times 10^8}{49.1 \times 10^{-12}} = 6.11 \times 10^{18}\) Hz.

Step 2: Plot the Square Root of Frequency vs Atomic Number

For each element, calculate the square root of the frequency \( \sqrt{u} \) and plot it against the atomic number of the element. Use the calculated frequencies from step 1:- Neon (Atomic number: 10): \(\sqrt{u} = \sqrt{2.05 \times 10^{17}}\)- Calcium (Atomic number: 20): \(\sqrt{u} = \sqrt{8.93 \times 10^{17}}\)- Zinc (Atomic number: 30): \(\sqrt{u} = \sqrt{2.09 \times 10^{18}}\)- Zirconium (Atomic number: 40): \(\sqrt{u} = \sqrt{3.82 \times 10^{18}}\)- Tin (Atomic number: 50): \(\sqrt{u} = \sqrt{6.11 \times 10^{18}}\)The plot will show a linear relationship between \(\sqrt{u}\) and the atomic number.

Step 3: Observation and Moseley's Law

The plot from Step 2 should exhibit a linear relationship between \(\sqrt{u}\) and the atomic number. Moseley discovered that the frequency of X-rays emitted is related to the atomic number by a linear relationship, \(\sqrt{u} = a(Z - b)\), where \(a\) and \(b\) are constants derived from experimental data. This linear trend allowed for the determination of atomic numbers of elements and prediction of missing elements.

Step 4: Predict Wavelength for Iron

Use the plot or linear relation between \(\sqrt{u}\) and atomic number to find the frequency for iron (Fe, atomic number 26) and then convert it into wavelength. Approximate the equation of the line based on linear regression:Suppose \(\sqrt{u} = mZ + c\). Plug in the atomic number for iron to find \(u\). Use \(u = \frac{c}{\lambda}\) to solve for \(\lambda\), where \(c = 3.00 \times 10^8\) m/s.

Step 5: Identify Element by Wavelength

For the given wavelength of 98.0 pm, calculate the corresponding frequency \(u = \frac{3.00 \times 10^8}{98.0 \times 10^{-12}} = 3.06 \times 10^{18}\) Hz. Find the square root, \(\sqrt{u} = \sqrt{3.06 \times 10^{18}}\), and check where this value fits on the plot from Step 2. This frequency suggests the element is close to Zirconium, so the element is likely Palladium (Pd), with an atomic number similar to Zr.

Key Concepts

Atomic Number

The atomic number is fundamental to understanding chemistry and physics. It represents the number of protons in the nucleus of an atom, distinct for each element. This characteristic defines the identity of an element and dictates its place on the periodic table.
For example, hydrogen has an atomic number of 1, helium is 2, and so forth. The higher the atomic number, the more complex the element's electron structure.

• It determines the element's chemical properties.
• It establishes the sequence of elements in the periodic table.

Moseley's work on X-rays solidified the concept of atomic numbers by showing that X-ray frequencies correlate directly with these numbers, effectively disproving the older belief that mass determined an element's identity.

X-ray Frequency

X-ray frequency refers to the number of oscillations of X-ray waves per second, typically measured in hertz (Hz). These high-energy electromagnetic waves can be used to explore the structure and properties of atoms.
The frequency (u) of X-rays is inversely proportional to the wavelength (1), represented by the formula:

• u = \frac{c}{1}, where \(c\) is the speed of light.

The elements emit X-rays at a unique frequency when excited, making it possible to use these frequencies to study atomic structure.

• Frequency increases with a decrease in wavelength.
• Higher atomic numbers tend to emit X-rays at higher frequencies.

This relationship allows scientists to determine elements' identity and characteristics without disturbing their physical structure.

Element Identification

Element identification through X-ray frequencies is a transformative method in modern science. By examining X-ray frequencies, scientists can identify unknown elements and their atomic numbers. Moseley's law introduced the idea that the square root of the frequency of X-rays is linear with the atomic number.

• Square Root of Frequency: \(\sqrt{u}\)
• Atomic Number: \(Z\)

This relationship permits not just identification but also the ordering of elements in the periodic table. When an element's X-ray frequency is known, it can be matched with others on a plot of \(\sqrt{u}\) vs. \(Z\). New or missing elements can then be predicted, as Moseley did in his time, forever altering our understanding and categorization of elements.

Wavelength Calculation

Calculating wavelength from frequency is crucial in exploring elements through X-ray spectroscopy. With the frequency of an element's X-rays, the wavelength can be computed using the speed of light. The formula involved is:

• \(u = \frac{c}{1}\), where \(c\) is the speed of light.

Wavelengths are frequently measured in picometers (pm), especially in the context of X-rays. These calculations are useful for examining and predicting other physical properties of elements.

• A decrease in wavelength signifies an increase in frequency, indicating higher energy X-rays.
• Through linear equations and plots, such as the one used by Moseley, wavelengths for unmeasured elements can be inferred.

Moseley's innovative use of wavelength calculations in discovering new elements underscores the power of this approach, prompting predictions about what elements should exist even before their discovery.