Question

A recently 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

The nuclear reaction representing the synthesis of bohrium is: \(^{249}_{97}Bk + ^{22}_{10}Ne \to ^{267}_{107}Bh + ^4_2He\) It would take approximately 55.4 seconds for the number of bohrium-267 atoms to decrease from 199 to 11. The electron configuration of elemental bohrium is: \[ [Rn]\ 5f^{14}\ 6d^{5}\ 7s^{2} \]

Step-by-step solution

Writing the nuclear reaction

To write the nuclear reaction, we can use the following notation: $$ ^{_A}{_Z}X + ^{_B}{_Y}W \to ^{_{A+B}}{_{Z+Y}}V + Q, $$ where ('\(_X\)' and '\(_W\)') are the reactants, ('\(_V\)') is the product, and ('\(_Q\)') represents any other particles released during the reaction. Using the atomic numbers (Z) and mass numbers (A) of berkelium-249 (Bk) and neon-22 (Ne), we can write the nuclear reaction representing the synthesis process: $$ ^{249}_{97}Bk + ^{22}_{10}Ne \to ^{267}_{107}Bh + X $$ Now, we need to find what \(X\) is. The atomic number must be conserved, so we have: $$Z_{X} = 97 + 10 - 107 = 0.$$ The mass number must also be conserved: $$A_{X} = 249 + 22 - 267 = 4.$$ This means the particle is an alpha particle, which we can denote by "\( ^4_2He \)": $$ ^{249}_{97}Bk + ^{22}_{10}Ne \to ^{267}_{107}Bh + ^4_2He $$

Calculating the time for Bh-267 atoms to decrease

We are given that the half-life of bohrium-267 is 15 seconds. We need to find the time it takes for the number of bohrium atoms to decrease from 199 to 11. We can use the half-life formula: $$N = N_0 \cdot \left(\frac{1}{2}\right)^{t/T_{1/2}}, $$ where '\(N\)' is the final number of atoms, '\(N_0\)' is the initial number of atoms, '\(t\)' is the time elapsed, and '\(T_{1/2}\)' is the half-life. Plugging in the values, we get: $$11 = 199 \cdot \left(\frac{1}{2}\right)^{t/15}.$$ To solve for '\(t\)', we will first divide both sides by 199: $$\frac{11}{199} = \left(\frac{1}{2}\right)^{t/15}.$$ Next, we will take the logarithm base 2 of both sides: $$\log_2{\frac{11}{199}} = \frac{t}{15}.$$ Now, we can solve for '\(t\)': $$t = 15 \cdot \log_2{\frac{11}{199}} \approx 55.4 \text{ seconds}.$$ So, it would take approximately 55.4 seconds for the number of bohrium-267 atoms to decrease from 199 to 11.

Determining the electron configuration of Bh

To determine the electron configuration of elemental bohrium (Bh), we need to know its atomic number, which is 107. The electron configuration can be obtained by following the pattern of the periodic table and filling up electron orbitals in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ... Since the atomic number of Bh is 107, its electron configuration should account for the 107 electrons. Following the pattern, we get: $$[Rn]\ 5f^{14}\ 6d^{5}\ 7s^{2},$$ where '\([Rn]\)' represents the noble gas core of radon (86 electrons). Elemental bohrium has the electron configuration \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\ 6s^2\ 4f^{14}\ 5d^{10}\ 6p^6\ 7s^2\ 5f^{14}\ 6d^{5}\).

Key concepts

Transuranium Elements

Transuranium elements are fascinating and unique. They lie beyond uranium on the periodic table and have atomic numbers greater than 92. These elements are mostly synthetic, created in laboratories through nuclear reactions. The process of creating new transuranium elements involves bombarding lighter elements with particles to create heavier elements. In our example, bohrium-267 was synthesized by bombarding berkelium-249 with neon-22. Transuranium elements tend to be unstable and exhibit radioactivity, meaning they decay into smaller, more stable atoms over time. This decay process offers exciting insights into nuclear physics and chemistry. Scientists study the reactions and properties of these elements to understand nuclear forces and the structure of matter. This area of research has implications for energy production and medical applications, among others.

Half-Life Calculations

Understanding half-life is crucial for studying radioactive elements like transuranium ones. Half-life is the time it takes for half of the atoms in a sample to decay. It’s a measure of the stability and rate at which the substance loses its radioactive properties.

• The half-life of bohrium-267 is 15 seconds, indicating rapid decay.
• To calculate how long it would take for the number of atoms to decrease from an initial amount to a smaller amount, we use the half-life formula:

\[ N = N_0 \cdot \left( \frac{1}{2} \right)^{t/T_{1/2}} \] Where:

• \( N \) = final number of atoms
• \( N_0 \) = initial number of atoms
• \( t \) = time elapsed
• \( T_{1/2} \) = half-life

For example, if we start with 199 bohrium-267 atoms, and we want to know when there will be just 11 left, we solve for \( t \). By rearranging and solving this equation, as demonstrated in our exercise, we found it to be approximately 55.4 seconds. This precise measure helps us predict the behavior and decay of radioactive elements.

Electron Configuration

Understanding electron configuration is key in determining how elements interact with each other. It describes the way in which electrons are distributed across the different atomic orbitals.
The electron configuration of an element is determined by following the order of filling orbital levels, which starts from the lowest energy orbitals and works up: 1s, 2s, 2p, 3s, and so on. For bohrium, with an atomic number of 107, the electron configuration is quite complex:

• The electron configuration is expressed as \[[Rn]5f^{14}6d^{5}7s^{2}\]
• This configuration indicates that bohrium's electrons fill up to the 6d and 7s subshells.
• The notation \([Rn]\) refers to the radon core, which accounts for the first 86 electrons.

Calculating electron configurations helps us understand the chemical properties and bonding behavior of elements, providing valuable information for studying both synthetic and naturally occurring elements.