Complete the equations for the following nuclear processes:
(a)`._(17)^(35) Cl + ._(0)^(1)n rarr... + ._(2)^(4)He`
(b) `._(92)^(235)U + ._(0)^(1) n rarr ...+ ._(54)^(137)Xe + 2 _(0)^(1)n`
(c) `._(13)^(27) Al + ._(2)^(4) He rarr ... + ._(0)^(1) n`
(d) `...(n,p) ._(16)^(35) S`
(e) `._(94)^(239) Pu (alpha, beta^(-))...`

Explanation

Let's solve the given nuclear processes step by step.

### (a)
**Given Reaction:**
\[ _{17}^{35} \text{Cl} + _{0}^{1} \text{n} \rightarrow ... + _{2}^{4} \text{He} \]

**Step 1:** Identify the atomic numbers and mass numbers.
- Chlorine (Cl) has atomic number 17 and mass number 35.
- Neutron (n) has atomic number 0 and mass number 1.
- Helium (He) has atomic number 2 and mass number 4.

**Step 2:** Set up the equation for atomic numbers.
\[ 17 + 0 = x + 2 \implies x = 15 \]

**Step 3:** Set up the equation for mass numbers.
\[ 35 + 1 = y + 4 \implies y = 32 \]

**Step 4:** Identify the element with atomic number 15 and mass number 32.
- This element is Phosphorus (P).

**Final Equation:**
\[ _{17}^{35} \text{Cl} + _{0}^{1} \text{n} \rightarrow _{15}^{32} \text{P} + _{2}^{4} \text{He} \]

### (b)
**Given Reaction:**
\[ _{92}^{235} \text{U} + _{0}^{1} \text{n} \rightarrow ... + _{54}^{137} \text{Xe} + 2 \, _{0}^{1} \text{n} \]

**Step 1:** Set up the equation for atomic numbers.
\[ 92 + 0 = x + 54 + 0 + 0 \implies x = 38 \]

**Step 2:** Set up the equation for mass numbers.
\[ 235 + 1 = y + 137 + 2 \implies y = 97 \]

**Step 3:** Identify the element with atomic number 38 and mass number 97.
- This element is Strontium (Sr).

**Final Equation:**
\[ _{92}^{235} \text{U} + _{0}^{1} \text{n} \rightarrow _{38}^{97} \text{Sr} + _{54}^{137} \text{Xe} + 2 \, _{0}^{1} \text{n} \]

### (c)
**Given Reaction:**
\[ _{13}^{27} \text{Al} + _{2}^{4} \text{He} \rightarrow ... + _{0}^{1} \text{n} \]

**Step 1:** Set up the equation for atomic numbers.
\[ 13 + 2 = x + 0 \implies x = 15 \]

**Step 2:** Set up the equation for mass numbers.
\[ 27 + 4 = y + 1 \implies y = 30 \]

**Step 3:** Identify the element with atomic number 15 and mass number 30.
- This element is Phosphorus (P).

**Final Equation:**
\[ _{13}^{27} \text{Al} + _{2}^{4} \text{He} \rightarrow _{15}^{30} \text{P} + _{0}^{1} \text{n} \]

### (d)
**Given Reaction:**
\[ ... + (n,p) \rightarrow _{16}^{35} \text{S} \]

**Step 1:** Set up the equation for atomic numbers.
- Let the initial atomic number be \( x \).
\[ x + 0 = 16 + 1 \implies x = 17 \]

**Step 2:** Set up the equation for mass numbers.
- Let the initial mass number be \( y \).
\[ y + 1 = 35 + 1 \implies y = 36 \]

**Step 3:** Identify the element with atomic number 17 and mass number 36.
- This element is Chlorine (Cl).

**Final Equation:**
\[ _{17}^{36} \text{Cl} + _{0}^{1} \text{n} \rightarrow _{16}^{35} \text{S} + _{1}^{1} \text{p} \]

### (e)
**Given Reaction:**
\[ _{94}^{239} \text{Pu} \, (\alpha, \beta^{-}) \rightarrow ... \]

**Step 1:** Set up the equation for atomic numbers.
\[ 94 + 2 - 1 = x \implies x = 95 \]

**Step 2:** Set up the equation for mass numbers.
\[ 239 + 4 + 0 = y \implies y = 243 \]

**Step 3:** Identify the element with atomic number 95 and mass number 243.
- This element is Americium (Am).

**Final Equation:**
\[ _{94}^{239} \text{Pu} \rightarrow _{95}^{243} \text{Am} + _{2}^{4} \text{He} + \text{(other products)} \]

### Summary of Final Reactions:
(a) \( _{17}^{35} \text{Cl} + _{0}^{1} \text{n} \rightarrow _{15}^{32} \text{P} + _{2}^{4} \text{He} \)
(b) \( _{92}^{235} \text{U} + _{0}^{1} \text{n} \rightarrow _{38}^{97} \text{Sr} + _{54}^{137} \text{Xe} + 2 \, _{0}^{1} \text{n} \)
(c) \( _{13}^{27} \text{Al} + _{2}^{4} \text{He} \rightarrow _{15}^{30} \text{P} + _{0}^{1} \text{n} \)
(d) \( _{17}^{36} \text{Cl} + _{0}^{1} \text{n} \rightarrow _{16}^{35} \text{S} + _{1}^{1} \text{p} \)
(e) \( _{94}^{239} \text{Pu} \rightarrow _{95}^{243} \text{Am} + _{2}^{4} \text{He} + \text{(other products)} \)