`._(92)U^(238)` emits `8 alpha-` particles and `6 beta-` particles. The `n//p` ratio in the product nucleus is

Options

A. `(62)/(41)`
B. `(60)/(41)`
C. `(61)/(42)`
D. `(62)/(42)`

Explanation

To solve the problem, we need to determine the neutron-to-proton (n/p) ratio in the product nucleus after the emission of 8 alpha particles and 6 beta particles from uranium-238.

### Step-by-Step Solution:

1. **Identify the Initial Nucleus:**
The initial nucleus is Uranium-238, denoted as \(_{92}^{238}\text{U}\). This means it has:
- Atomic number (Z) = 92 (number of protons)
- Mass number (A) = 238 (total number of protons and neutrons)

2. **Calculate the Effect of Alpha Particle Emission:**
Each alpha particle emission reduces the mass number by 4 and the atomic number by 2. Since 8 alpha particles are emitted:
- Change in mass number = \(8 \times 4 = 32\)
- Change in atomic number = \(8 \times 2 = 16\)

Therefore, after emitting 8 alpha particles:
- New mass number = \(238 - 32 = 206\)
- New atomic number = \(92 - 16 = 76\)

The product nucleus after alpha emission is \(_{76}^{206}\text{X}\).

3. **Calculate the Effect of Beta Particle Emission:**
Each beta particle emission increases the atomic number by 1 while the mass number remains unchanged. Since 6 beta particles are emitted:
- Change in atomic number = \(6 \times 1 = 6\)

Therefore, after emitting 6 beta particles from the nucleus formed after alpha emission:
- New atomic number = \(76 + 6 = 82\)
- Mass number remains = 206

The final product nucleus after both emissions is \(_{82}^{206}\text{Y}\).

4. **Calculate Neutrons and Protons in the Final Nucleus:**
- Number of protons (Z) = 82
- Number of neutrons (N) can be calculated using the formula:
\[ N = A - Z = 206 - 82 = 124 \]

5. **Calculate the Neutron-to-Proton Ratio:**
The neutron-to-proton ratio (n/p) is given by:
\[ \text{n/p} = \frac{N}{Z} = \frac{124}{82} \]

This can be simplified:
\[ \text{n/p} = \frac{62}{41} \]

6. **Final Answer:**
The neutron-to-proton ratio in the product nucleus is \(62:41\).