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?

Short Answer

Expert verified
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.

Step by step solution

01

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: \[_{97}^{249}\textrm{Bk} + _{10}^{22}\textrm{Ne} \rightarrow _{107}^{267}\textrm{Bh}\]
02

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
03

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:
04

Nuclear Reaction:

\[_{97}^{249}\textrm{Bk} + _{10}^{22}\textrm{Ne} \rightarrow _{107}^{267}\textrm{Bh}\]
05

Time Elapsed:

\(n = \log_{\frac{1}{2}}{\frac{11}{199}}≈4.2\) Elapsed Time ≈ 4.2 × 15.0 seconds ≈ 63.0 seconds
06

Electron Configuration:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d⁵

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