Problem 95
Question
A reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is 15.0 seconds. If 199 atoms of bohrium-267 could be synthesized, how much time would elapse before only 11 atoms of bohrium-267 remain? What is the expected electron configuration of elemental bohrium?
Step-by-Step Solution
Verified Answer
The nuclear reaction for the synthesis of bohrium-267 is:
\[_{97}^{249}\textrm{Bk} + _{10}^{22}\textrm{Ne} \rightarrow _{107}^{267}\textrm{Bh}\]
The time elapsed before only 11 atoms of bohrium-267 remain is approximately 63.0 seconds.
The expected electron configuration of elemental bohrium is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d⁵.
1Step 1: 1. Writing the Nuclear Reaction
To write the nuclear reaction for the synthesis of bohrium-267, we need to combine berkelium-249 and neon-22. The equation should follow the form:
\(A(a, b)C\), where A and C are reactants, and a and b are the products of the reaction. In this case, A is Berkelium-249, a is Neon-22, and C is Bohrium-267. The reaction would look like this:
\[_{97}^{249}\textrm{Bk} + _{10}^{22}\textrm{Ne} \rightarrow _{107}^{267}\textrm{Bh}\]
2Step 2: 2. Calculating the Time Elapsed
We will use the concept of half-life to determine the time elapsed before only 11 atoms of bohrium-267 remain. First, we need to find the number of half-lives that need to pass for only 11 atoms to remain from the initial 199 atoms:
Number of remaining atoms = Initial number of atoms × left-over percentage
Using the formula for half-life, we get:
\(11 = 199 × \left(\frac{1}{2}\right)^n\)
Where n is the number of half-lives. Solve for n:
\(n = \log_{\frac{1}{2}}{\frac{11}{199}}\)
Now, we can find the time elapsed by multiplying the number of half-lives by the half-life of bohrium-267:
Elapsed time = n × half-life
3Step 3: 3. Electron Configuration of Elemental Bohrium
The electron configuration of an element is determined by its atomic number. Elemental bohrium has an atomic number of 107, so we need to fill up its orbitals with 107 electrons following the Aufbau principle. The electron configuration for bohrium would be:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d⁵
So, finally:
4Step 4: Nuclear Reaction:
\[_{97}^{249}\textrm{Bk} + _{10}^{22}\textrm{Ne} \rightarrow _{107}^{267}\textrm{Bh}\]
5Step 5: Time Elapsed:
\(n = \log_{\frac{1}{2}}{\frac{11}{199}}≈4.2\)
Elapsed Time ≈ 4.2 × 15.0 seconds ≈ 63.0 seconds
6Step 6: Electron Configuration:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d⁵
Key Concepts
Transuranium ElementsHalf-Life CalculationElectron Configuration
Transuranium Elements
Transuranium elements are synthetic elements with atomic numbers greater than that of uranium (92). These elements do not occur naturally and are created in laboratories. One exciting aspect of transuranium elements is that they offer a glimpse into the behavior of atomic nuclei at extreme conditions. These elements, such as bohrium (Bh) with atomic number 107, are produced through nuclear reactions. A typical reaction involves bombarding a target nucleus with a projectile, like combining berkelium-249 with neon-22 to create bohrium-267.
- Synthesis: The process of creating transuranium elements usually involves particle accelerators that propel ions to hit target materials, forming new elements.
- Applications: While some transuranium elements are primarily of scientific interest, they can provide valuable insights into the forces holding atomic nuclei together.
Half-Life Calculation
The concept of half-life is key to understanding how long a radioactive element remains active. Half-life is the time required for half of the radioactive atoms in a sample to decay. For bohrium-267, the half-life is remarkably short at just 15 seconds. This means that after 15 seconds, only half of any sample of bohrium-267 remains. Calculating elapsed time using this concept involves determining how many half-lives pass for a given reduction in atoms.
- Formula: The formula for the number of half-lives is given by = \(\log_{\frac{1}{2}}(\frac{N}{N_0})\), where \(N_0\) is the initial quantity, and \(N\) is the remaining quantity.
- Example: For bohrium-267, determining the time for reducing to 11 atoms from 199 involves solving \(11 = 199 \times (\frac{1}{2})^n\).
- Time Calculation: With \(n \approx 4.2\), multiply by the half-life of 15 seconds to find the elapsed time of approximately 63 seconds.
Electron Configuration
Electron configuration describes how electrons are distributed in an atom's orbitals, following rules based on quantum mechanics. For bohrium, which has the atomic number 107, understanding its electron configuration requires filling its orbitals with 107 electrons, according to the Aufbau principle, which tells us the order in which orbitals are filled.
- Rules: Electrons fill the lowest energy levels first before moving to higher ones. The Aufbau principle outlines this order.
- Configuration: Bohrium's configuration is complex, reflecting periodic table trends. It is written as 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d⁵.
- Utility: Knowing electron configurations helps in understanding chemical reactivity, bonding, and properties of elements.
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