Question

Complete and balance each of the following nuclear equations: a. \(\mathrm{Bi}-209+\mathrm{Cr}-54 \longrightarrow ?+{ }_{0}^{1} n\) b. Np-237 undergoes alpha decay. c. \({ }_{2}^{4} \mathrm{He}+{ }_{95}^{241} \mathrm{Am} \longrightarrow ?+2_{0}^{1} n\)

Short answer

a. \( _{107}^{262} \text{Bh} \rightarrow ? + _0^1 n \). b. \( \text{Np}-237 \rightarrow _{91}^{233} \text{Pa} + _2^4 \text{He} \). c. \( _2^4 \text{He} + _{95}^{241} \text{Am} \rightarrow _{97}^{243} \text{Bk} + 2_0^1 n \).

Step-by-step solution

Identify the known particles and isotopes

For each equation, identify the given isotopes and particles. This includes their mass numbers (upper value) and atomic numbers (lower value).

Determine the missing particle or isotope (Part a)

In the equation \( \text{Bi}-209 + \text{Cr}-54 \rightarrow ? + _0^1 n \) : First, add the mass numbers of Bi-209 and Cr-54: \[ 209 + 54 = 263 \]. Then, add the atomic numbers: \[ 83 + 24 = 107 \]. The product side already has a neutron \( (_0^1 n) \), so the missing isotope must have a mass number of 263 - 1 = 262 and an atomic number of 107. Identify the element with atomic number 107, which is Bohrium (Bh).

Complete and balance the equation (Part a)

The balanced equation is \( \text{Bi}-209 + \text{Cr}-54 \rightarrow _{107}^{262} \text{Bh} + _0^1 n \).

Determine the particles emitted in alpha decay (Part b)

During alpha decay, an isotope emits an alpha particle \( (_2^4 \text{He}) \). The equation for Np-237 undergoing alpha decay is: \( \text{Np}-237 \rightarrow ? + _2^4 \text{He} \).

Determine the daughter nucleus (Part b)

First, subtract the mass number of the alpha particle from Np-237: \[ 237 - 4 = 233 \]. Then subtract the atomic number: \[ 93 - 2 = 91 \]. Find the element with atomic number 91, which is Protactinium (Pa). The equation is: \( \text{Np}-237 \rightarrow _{91}^{233} \text{Pa} + _2^4 \text{He} \).

Identify the known particles and isotopes (Part c)

In the equation \( _2^4 \text{He} + _{95}^{241} \text{Am} \rightarrow ? + 2_0^1 n \), start by determining the total mass number and atomic number. The mass number is: \[ 4 + 241 = 245 \], and the atomic number is: \[ 2 + 95 = 97 \].

Determine the missing particle or isotope (Part c)

Considering the two neutrons on the right, subtract their mass numbers from the total: \[ 245 - 2 \times 1 = 243 \]. The atomic number remains 97. Identify the element with atomic number 97, which is Berkelium (Bk). The balanced equation is: \[ _2^4 \text{He} + _{95}^{241} \text{Am} \rightarrow _{97}^{243} \text{Bk} + 2_0^1 n \].

Key concepts

Nuclear Chemistry

Nuclear chemistry focuses on the reactions and properties of atomic nuclei. It's a branch of chemistry associated with changes in the nucleus, in contrast to the outer electrons studied in traditional chemistry. Nuclear reactions involve interactions between the protons and neutrons in the nucleus.

Key processes in nuclear chemistry include:

• Fission: Splitting a heavy nucleus into smaller nuclei, releasing energy.
• Fusion: Combining light nuclei into a heavier nucleus, releasing even more energy than fission.
• Radioactive Decay: Emission of particles or radiation as unstable nuclei become more stable.

Understanding these reactions is crucial for various applications, from nuclear power generation to medical treatments.

Alpha Decay

Alpha decay is a process where an unstable nucleus emits an alpha particle, becoming a different element. An alpha particle consists of two protons and two neutrons—essentially a helium-4 nucleus. This emission decreases the mass number by 4 and the atomic number by 2.
For example, consider Np-237 undergoing alpha decay:

• First, subtract the mass number of the alpha particle from Np-237: \(237 - 4 = 233\).
• Then subtract the atomic number: \(93 - 2 = 91\).
• Identify the element with atomic number 91, which is Protactinium (Pa).

The balanced equation would be:
\[ \text{Np}-237 \rightarrow _{91}^{233} \text{Pa} + _2^4 \text{He} \]

Isotopes

Isotopes are varieties of the same element, differing in the number of neutrons. Though they have the same atomic number (number of protons), their different mass numbers (sum of protons and neutrons) attribute unique properties.
Important points about isotopes:

• Isotopes can be stable or unstable (radioactive).
• Unstable isotopes often undergo radioactive decay to achieve stability.
• Applications of isotopes include medical imaging, radiotherapy, and carbon dating.

Isotopes play a significant role in nuclear chemistry, influencing the outcomes and balance of nuclear equations.

Neutron Emission

Neutron emission occurs when a nucleus releases a neutron, often during nuclear reactions or radioactive decay. Neutrons are neutral particles, so their emission does not change the atomic number but decreases the mass number by one.
For instance, in the reaction \( \text{Bi}-209 + \text{Cr}-54 \rightarrow _{107}^{262} \text{Bh} + _0^1 n \):

• The neutron emitted is represented as \( _0^1 n \), with a mass number of 1 and no charge.
• The mass number of the product Bohrium (Bh) is calculated by subtracting the neutron's mass number from the total: \(263 - 1 = 262 \).
• The atomic number of Bohrium remains 107.

Neutron emission helps us understand how nuclei can be transformed and play a critical role in nuclear equations.