When `._(17)Cl^(35)` undergoes `(n,p)` reaction, the radioisotope formed is

Options

A. `._(15)P^(32)`
B. `._(16)S^(35)`
C. `._(16)S^(34)`
D. `._(15)P^(34)`

Answer

To solve the problem of what radioisotope is formed when \( \text{Cl}^{35}_{17} \) undergoes an \( (n,p) \) reaction, we can follow these steps:

### Step 1: Understand the \( (n,p) \) Reaction
The \( (n,p) \) reaction involves a neutron entering the nucleus of an atom and a proton being emitted. This means that:
- A neutron is added to the nucleus.
- A proton is removed from the nucleus.

### Step 2: Identify the Initial Nucleus
The initial nucleus is \( \text{Cl}^{35}_{17} \):
- Mass number (A) = 35
- Atomic number (Z) = 17

### Step 3: Determine the Changes in Mass and Atomic Numbers
In an \( (n,p) \) reaction:
- The mass number (A) remains unchanged because a neutron is added and a proton is removed, which does not affect the total count of nucleons.
- The atomic number (Z) decreases by 1 because a proton is emitted.

### Step 4: Calculate the New Atomic Number and Element
- New atomic number (Z) = 17 - 1 = 16
- The mass number (A) remains 35.

### Step 5: Identify the Element with Atomic Number 16
The element with atomic number 16 is sulfur (S). Therefore, the resulting radioisotope is \( \text{S}^{35}_{16} \).

### Final Answer
The radioisotope formed when \( \text{Cl}^{35}_{17} \) undergoes an \( (n,p) \) reaction is \( \text{S}^{35}_{16} \).

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