Question

What are the respective numbers of \(\alpha\) and \(\beta\) particles emitted respectively in the following radioactive decay? \[ {}^{200}_{90}X \rightarrow {}^{168}_{80}Y \]

• A. \(8\) and \(8\)
• B. \(8\) and \(6\)
• C. \(6\) and \(8\)
• D. \(6\) and \(6\)

Hint:

For \(\alpha\)-decay, mass number decreases by \(4\) and atomic number decreases by \(2\). For \(\beta^{-}\)-decay, mass number remains same and atomic number increases by \(1\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Radioactive decay involves the transformation of an unstable nucleus into a more stable one by emitting particles.
The two primary modes of decay discussed here are alpha (\(\alpha\)) and beta (\(\beta^-\)) decay.
An alpha particle is essentially a helium nucleus, represented as \({}_{2}^{4}He\).
When a nucleus emits an alpha particle, its mass number (\(A\)) decreases by 4 units because it loses two protons and two neutrons.
Simultaneously, its atomic number (\(Z\)) decreases by 2 units because it loses two protons.
A beta-minus (\(\beta^-\)) particle is a high-speed electron emitted from the nucleus, represented as \({}_{-1}^{0}e\).
During beta decay, a neutron inside the nucleus transforms into a proton, an electron, and an antineutrino.
Because a neutron is replaced by a proton, the total number of nucleons (mass number \(A\)) remains the same.
However, because there is now one additional proton, the atomic number (\(Z\)) increases by 1.
Step 2: Key Formula or Approach:
Let \(n_{\alpha}\) be the total number of alpha particles emitted and \(n_{\beta}\) be the total number of beta particles emitted.
The total change in the mass number is solely due to alpha particles:
\[ \Delta A = A_{initial} - A_{final} = 4 \times n_{\alpha} \]
The total change in the atomic number is the combined effect of both:
\[ Z_{final} = Z_{initial} - 2n_{\alpha} + 1n_{\beta} \]
Step 3: Detailed Explanation:
Given the decay: \({}_{90}^{200} X \rightarrow {}_{80}^{168} Y\).
Initial values: \(A_1 = 200\), \(Z_1 = 90\).
Final values: \(A_2 = 168\), \(Z_2 = 80\).

First, we analyze the change in the mass number to find the number of alpha particles.
Since beta decay does not change the mass number, the entire loss of 32 units (\(200 - 168 = 32\)) must come from alpha emissions.
As each alpha particle accounts for a loss of 4 units:
\[ 4 \times n_{\alpha} = 32 \]
\[ n_{\alpha} = \frac{32}{4} = 8 \]
So, 8 alpha particles are emitted.

Next, we calculate the expected atomic number after these 8 alpha emissions.
Each alpha particle reduces the atomic number by 2. For 8 particles, the reduction is \(8 \times 2 = 16\).
\[ Z_{inter} = 90 - 16 = 74 \]
However, the final atomic number of the daughter nucleus \(Y\) is given as 80.
The difference between the actual atomic number (80) and the intermediate atomic number (74) is caused by beta particles.
Each beta particle increases the atomic number by 1.
\[ 80 = 74 + n_{\beta} \]
\[ n_{\beta} = 80 - 74 = 6 \]
Thus, the decay involves the emission of 8 alpha particles and 6 beta particles.
Step 4: Final Answer:
The numbers of \(\alpha\) and \(\beta\) particles are 8 and 6, which matches option (B).