Sulphur `35`(34,96903 amu) emits a `beta-` particles but no `gamma`-rays. The product is c hlorine `-35` (`34,96885` amu),. The maximum energy carried by `beta-` particle is:

Options

A. 16.758 MeV
B. 1.6785 MeV
C. 0.16758 MeV
D. 0.016758 MeV

Explanation

To find the maximum energy carried by the beta particle emitted during the decay of Sulfur-35 to Chlorine-35, we can follow these steps:

### Step 1: Identify the masses involved
We have the following masses:
- Mass of Sulfur-35 (S-35): 34.96903 amu
- Mass of Chlorine-35 (Cl-35): 34.96885 amu

### Step 2: Calculate the mass defect
The mass defect is the difference between the mass of the parent nucleus (Sulfur-35) and the mass of the daughter nucleus (Chlorine-35).

\[ \text{Mass defect} = \text{Mass of S-35} - \text{Mass of Cl-35} \]

Substituting the values:

\[ \text{Mass defect} = 34.96903 \, \text{amu} - 34.96885 \, \text{amu} = 0.00018 \, \text{amu} \]

### Step 3: Convert mass defect to energy
To convert the mass defect into energy, we use the formula:

\[ \text{Energy (MeV)} = \text{Mass defect (grams)} \times 931 \, \text{MeV} \]

First, we need to convert the mass defect from amu to grams. Since 1 amu is approximately \(1.660539 \times 10^{-24}\) grams, we convert:

\[ 0.00018 \, \text{amu} = 0.00018 \times 1.660539 \times 10^{-24} \, \text{grams} = 2.99 \times 10^{-28} \, \text{grams} \]

Now, substituting the mass defect into the energy formula:

\[ \text{Energy} = 0.00018 \, \text{amu} \times 931 \, \text{MeV} = 0.16758 \, \text{MeV} \]

### Step 4: Conclusion
The maximum energy carried by the beta particle emitted during the decay of Sulfur-35 to Chlorine-35 is approximately **0.1675 MeV**.

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