The number of `alpha` and `beta`- emitted during the radioactive decay chain starting from `._(226)^(88) Ra` and ending at `._(206)^(82) Pb` is

Options

A. `3alpha` and `6beta^(-)`
B. `4alpha` and `5beta^(-)`
C. `5alpha` and `4beta^(-)`
D. `6alpha` and `6beta^(-)`

Answer

To solve the problem of determining the number of alpha (α) and beta (β) particles emitted during the radioactive decay chain starting from \( _{88}^{226}Ra \) and ending at \( _{82}^{206}Pb \), we can follow these steps:

### Step 1: Understand the decay process
The decay process involves the emission of alpha and beta particles. An alpha particle (α) is a helium nucleus, which has 2 protons and 2 neutrons, while a beta particle (β) is an electron emitted during the decay of a neutron into a proton.

### Step 2: Write the decay equation
We can express the decay as:
\[ _{88}^{226}Ra \rightarrow N \, \alpha + M \, \beta + _{82}^{206}Pb \]
where \( N \) is the number of alpha particles emitted and \( M \) is the number of beta particles emitted.

### Step 3: Calculate the change in mass number (A)
The mass number (A) changes as follows:
- Initial mass number of Ra = 226
- Final mass number of Pb = 206

The change in mass number due to alpha decay is \( 4N \) (since each alpha particle has a mass number of 4). Therefore, we can write:
\[ 226 - 4N = 206 \]
Solving for \( N \):
\[ 4N = 226 - 206 = 20 \\ N = \frac{20}{4} = 5 \]

### Step 4: Calculate the change in atomic number (Z)
The atomic number (Z) changes as follows:
- Initial atomic number of Ra = 88
- Final atomic number of Pb = 82

The change in atomic number due to alpha decay is \( 2N \) and due to beta decay is \( M \). Therefore, we can write:
\[ 88 - 2N + M = 82 \]
Substituting \( N = 5 \):
\[ 88 - 2(5) + M = 82 \\ 88 - 10 + M = 82 \\ M = 82 - 78 = 4 \]

### Step 5: Conclusion
From the calculations, we find:
- Number of alpha particles (N) = 5
- Number of beta particles (M) = 4

Thus, the answer is that during the decay chain, **5 alpha particles and 4 beta particles are emitted**.

### Final Answer:
The number of alpha and beta particles emitted is:
- **5 alpha particles**
- **4 beta particles**