Step 1: Understanding the Concept:
Nuclear radioactivity involves the emission of particles to reach a state of lower energy and higher stability. When a nucleus undergoes alpha ($\alpha$) decay, it ejects a helium nucleus ($^{4}_{2}\text{He}$), which leads to a decrease in the total mass number ($A$) by 4 units and a decrease in the atomic number ($Z$) by 2 units. On the other hand, beta ($\beta^-$) decay involves the transformation of a neutron into a proton within the nucleus, accompanied by the emission of an electron ($^{0}_{-1}\text{e}$). This specific process results in no change to the mass number but increases the atomic number by 1 unit. By applying the laws of conservation of mass and charge, we can determine the specific sequence of emissions.
Step 2: Key Formula or Approach:
The overall decay process can be modeled using the following balance equation:
$^{200}_{90}\text{X} \rightarrow ^{168}_{80}\text{Y} + n_{\alpha}(^{4}_{2}\text{He}) + n_{\beta}(^{0}_{-1}\text{e})$
1. Conservation of Mass Number ($A$): $A_{initial} = A_{final} + 4(n_{\alpha}) + 0(n_{\beta})$
2. Conservation of Atomic Number ($Z$): $Z_{initial} = Z_{final} + 2(n_{\alpha}) - 1(n_{\beta})$
We solve for $n_{\alpha}$ first using the mass balance, and then solve for $n_{\beta}$ using the atomic number balance.
Step 3: Detailed Explanation:
We begin by analyzing the change in the mass number from the parent nucleus X to the daughter nucleus Y. The mass number drops from 200 to 168.
Since the emission of beta particles does not affect the mass number, the entire change of $200 - 168 = 32$ units must be caused by the loss of alpha particles.
Setting up the equation: $4 \times n_{\alpha} = 32$. Dividing both sides by 4 gives us $n_{\alpha} = 8$. This confirms that 8 alpha particles were emitted during the process.
Next, we look at the atomic number conservation. The initial atomic number is 90 and the final is 80.
The total reduction in atomic number due to 8 alpha particles is $8 \times 2 = 16$. This would theoretically result in an atomic number of $90 - 16 = 74$.
However, the actual final atomic number is 80. This means the atomic number must have increased by 6 units through beta decay ($80 - 74 = 6$).
Substituting into our formula: $90 = 80 + 2(8) - n_{\beta}$, which simplifies to $90 = 80 + 16 - n_{\beta}$.
Solving for $n_{\beta}$: $90 = 96 - n_{\beta} \implies n_{\beta} = 6$. Therefore, 6 beta particles were emitted.
Step 4: Final Answer:
The number of $\alpha$ and $\beta$ particles emitted are 8 and 6 respectively.