Which of the following notations shows the products incorrectly?
a. `._(96)Cm^(242) (alpha, 2n) ._(97)Bk^(243)` b. `._(5)B^(10) (alpha,n) ._(7)N^(13)`
c. `._(7)N^(14) (n,p) ._(6)C^(14)` d. `._(14)Si^(28) (d,n) ._(15)p^(29)`

Options

A. a. `._(96)Cm^(242) (alpha, 2n) ._(97)Bk^(243)`
B. b. `._(5)B^(10) (alpha,n) ._(7)N^(13)`
C. c. `._(7)N^(14) (n,p) ._(6)C^(14)`
D. d. `._(14)Si^(28) (d,n) ._(15)p^(29)`

Explanation

To solve the question of which notation shows the products incorrectly, we will analyze each option step by step.

### Step 1: Analyze Option A
**Notation:** `._(96)Cm^(242) (alpha, 2n) ._(97)Bk^(243)`

- **Initial Species:** Curium-242 (Cm with atomic number 96)
- **Bombarding Particle:** Alpha particle (He-4, atomic number 2)
- **Products:** Bk-243 (atomic number 97) and 2 neutrons (2n)

**Mass Number Calculation:**
- Left Side: 242 (Cm) + 4 (alpha particle) = 246
- Right Side: 243 (Bk) + 2 (neutrons) = 245

**Result:** 246 ≠ 245 (not equal)

**Atomic Number Calculation:**
- Left Side: 96 (Cm) + 2 (alpha particle) = 98
- Right Side: 97 (Bk)

**Result:** 98 ≠ 97 (not equal)

**Conclusion:** This notation is incorrect.

### Step 2: Analyze Option B
**Notation:** `._(5)B^(10) (alpha,n) ._(7)N^(13)`

- **Initial Species:** Boron-10 (B with atomic number 5)
- **Bombarding Particle:** Alpha particle (He-4, atomic number 2)
- **Products:** Nitrogen-13 (atomic number 7) and 1 neutron (n)

**Mass Number Calculation:**
- Left Side: 10 (B) + 4 (alpha particle) = 14
- Right Side: 13 (N) + 1 (neutron) = 14

**Result:** 14 = 14 (equal)

**Atomic Number Calculation:**
- Left Side: 5 (B) + 2 (alpha particle) = 7
- Right Side: 7 (N)

**Result:** 7 = 7 (equal)

**Conclusion:** This notation is correct.

### Step 3: Analyze Option C
**Notation:** `._(7)N^(14) (n,p) ._(6)C^(14)`

- **Initial Species:** Nitrogen-14 (N with atomic number 7)
- **Bombarding Particle:** Neutron (n)
- **Products:** Carbon-14 (C with atomic number 6) and a proton (p)

**Mass Number Calculation:**
- Left Side: 14 (N) + 1 (neutron) = 15
- Right Side: 14 (C) + 1 (proton) = 15

**Result:** 15 = 15 (equal)

**Atomic Number Calculation:**
- Left Side: 7 (N) + 0 (neutron) = 7
- Right Side: 6 (C) + 1 (proton) = 7

**Result:** 7 = 7 (equal)

**Conclusion:** This notation is correct.

### Step 4: Analyze Option D
**Notation:** `._(14)Si^(28) (d,n) ._(15)p^(29)`

- **Initial Species:** Silicon-28 (Si with atomic number 14)
- **Bombarding Particle:** Deuterium (d, which is He-2)
- **Products:** Phosphorus-29 (P with atomic number 15) and a neutron (n)

**Mass Number Calculation:**
- Left Side: 28 (Si) + 2 (deuterium) = 30
- Right Side: 29 (P) + 1 (neutron) = 30

**Result:** 30 = 30 (equal)

**Atomic Number Calculation:**
- Left Side: 14 (Si) + 1 (deuterium) = 15
- Right Side: 15 (P)

**Result:** 15 = 15 (equal)

**Conclusion:** This notation is correct.

### Final Conclusion
The only incorrect notation is **Option A**.

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