For the following nuclear decay series segment,
$_{90}^{234}{Th}$ → → → $_{90}^{230}{Th}$
the overall emitted particles are

Question

For the following nuclear decay series segment,$_{90}^{234}{Th}$ &rarr; &rarr; &rarr; $_{90}^{230}{Th}$the overall emitted particles are

Prepp Team · Accepted Answer

Understanding Nuclear Decay Series and Emitted Particles

This question asks us to identify the particles emitted during a specific segment of a nuclear decay series, where Thorium-234 transforms into Thorium-230. We are given the initial nucleus, $_{90}^{234}\text{Th}$, and the final nucleus, $_{90}^{230}\text{Th}$. To find the emitted particles, we need to analyze the changes in the mass number and the atomic number from the start of the segment to the end.

Analyzing Changes in Mass Number and Atomic Number

The initial nucleus is $_{90}^{234}\text{Th}$. Here, the mass number (superscript) is 234, and the atomic number (subscript) is 90.

The final nucleus is $_{90}^{230}\text{Th}$. Here, the mass number is 230, and the atomic number is 90.

Change in mass number = Final mass number - Initial mass number = $230 - 234 = -4$.
Change in atomic number = Final atomic number - Initial atomic number = $90 - 90 = 0$.

The total decay process results in a decrease of 4 in the mass number and no change in the atomic number.

Properties of Common Nuclear Decay Particles

Let's consider the common types of particles emitted during nuclear decay and how they affect the mass number and atomic number:

Alpha particle ($\alpha$): A helium nucleus $_{2}^{4}\text{He}$. Emission decreases the mass number by 4 and the atomic number by 2.
Beta-minus particle ($\beta^-$): An electron $_{-1}^{0}\text{e}$. Emission keeps the mass number the same (change of 0) and increases the atomic number by 1.
Neutron ($n$): A neutron $_{0}^{1}\text{n}$. Emission decreases the mass number by 1 and keeps the atomic number the same (change of 0).

Balancing Nuclear Equations to Find Emitted Particles

Let's assume that the decay involves $x$ alpha particles and $y$ beta-minus particles. We can write a general equation for the transformation:

\[_{90}^{234}\text{Th} \rightarrow x \cdot _{2}^{4}\text{He} + y \cdot _{-1}^{0}\text{e} + _{90}^{230}\text{Th}\]

Now, we balance the mass numbers (superscripts) on both sides of the equation:

\[234 = x \cdot 4 + y \cdot 0 + 230\]

\[234 = 4x + 230\]

Subtract 230 from both sides:

\[234 - 230 = 4x\]

\[4 = 4x\]

\[x = \frac{4}{4} = 1\]

So, there is 1 alpha particle emitted.

Next, we balance the atomic numbers (subscripts) on both sides of the equation:

\[90 = x \cdot 2 + y \cdot (-1) + 90\]

Substitute the value of $x=1$ into the equation:

\[90 = 1 \cdot 2 + y \cdot (-1) + 90\]

\[90 = 2 - y + 90\]

Subtract 90 from both sides:

\[90 - 90 = 2 - y\]

\[0 = 2 - y\]

Add $y$ to both sides:

\[y = 2\]

So, there are 2 beta-minus particles emitted.

The emitted particles in this nuclear decay series segment are one alpha particle and two beta particles.

Evaluating the Options

Based on our analysis, the emitted particles are two $\beta$ and one $\alpha$.

Let's check the options:

one $\beta$, one $\alpha$, and one neutron: This doesn't match our result (two $\beta$, one $\alpha$).
two $\beta$ and one $\alpha$: This matches our result.
three $\beta$: This doesn't match our result.
two $\beta$ and one neutron: This doesn't match our result.

Our calculation confirms that the overall emitted particles are two beta particles and one alpha particle.

Particle	Symbol	Change in Mass Number	Change in Atomic Number
Alpha	$_{2}^{4}\text{He}$ or $\alpha$	-4	-2
Beta-minus	$_{-1}^{0}\text{e}$ or $\beta^-$	0	+1
Neutron	$_{0}^{1}\text{n}$ or n	-1	0

Revision Table: Nuclear Decay Calculations

Parameter	Initial ($_{90}^{234}\text{Th}$)	Final ($_{90}^{230}\text{Th}$)	Total Change
Mass Number	234	230	$230 - 234 = -4$
Atomic Number	90	90	$90 - 90 = 0$

Particle Count	Contribution to Mass Change	Contribution to Atomic Change	Calculation
$x$ Alpha	$x \times (-4)$	$x \times (-2)$	$x=1$ (from mass balance)
$y$ Beta-minus	$y \times (0)$	$y \times (+1)$	$y=2$ (from atomic balance using $x=1$)
Total Change	$-4x$	$-2x + y$	Must match total change (-4, 0)

Additional Information on Nuclear Decay

Nuclear decay is a process where an unstable atomic nucleus loses energy by emitting radiation. This radiation can be in the form of particles (like alpha or beta) or electromagnetic waves (like gamma rays). Radioactive decay occurs at a specific rate, often described by its half-life.

Nuclear decay series, like the one involving Thorium, are sequences of decays where a parent nucleus undergoes a series of transformations through alpha or beta emissions until it reaches a stable daughter nucleus. There are several naturally occurring decay series, originating from isotopes of Uranium, Thorium, and Actinium.

Balancing nuclear equations is crucial for understanding decay processes. It involves ensuring that the total mass number and the total atomic number are conserved before and after the decay. Gamma ($\gamma$) rays are high-energy photons and do not change the mass number or atomic number of the nucleus, so they are often omitted when balancing particle emissions but are important for energy considerations.

Prepp Team · Answer

one &beta;, one &alpha;, and one neutron

Prepp Team · Answer

three &beta;

Prepp Team · Answer

two &beta; and one netron

Particle Count	Contribution to Mass Change	Contribution to Atomic Change	Calculation
\(x\) Alpha	\(x \times (-4)\)	\(x \times (-2)\)	\(x=1\) (from mass balance)
\(y\) Beta-minus	\(y \times (0)\)	\(y \times (+1)\)	\(y=2\) (from atomic balance using \(x=1\))
Total Change	\(-4x\)	\(-2x + y\)	Must match total change (-4, 0)