Nuclear reaction equations demonstrating various radioactive decay and transmutation processes, with unknown components marked as X.

Let's solve each nuclear reaction step by step. In nuclear reactions, both mass number (A) and atomic number (Z) must be conserved.

We will use the following rules:

- The sum of mass numbers on both sides must be equal.
- The sum of atomic numbers on both sides must be equal.
- Use known particles:
- β⁻ (beta minus): \( ^0_{-1}\beta \) → electron
- α (alpha): \( ^4_2\alpha \) → helium nucleus
- n (neutron): \( ^1_0n \)
- p (proton): \( ^1_1p \)

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a) \( X \rightarrow {}^{97}_{41}\text{Nb} + {}^0_{-1}\beta \)

This is beta decay. A neutron turns into a proton, emitting an electron (\( \beta^- \)).

To find X, reverse the decay:
The parent nucleus has:
- Mass number: \( 97 + 0 = 97 \)
- Atomic number: \( 41 + (-1) = 40 \)

So, \( X = {}^{97}_{40}\text{Zr} \) (Zirconium)

✔ Answer: \( {}^{97}_{40}\text{Zr} \)

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b) \( {}^{205}_{83}\text{Bi} \rightarrow {}^{205}_{82}\text{Pb} + X \)

Mass: 205 → 205 + A ⇒ A = 0
Atomic number: 83 → 82 + Z ⇒ Z = +1

So, \( X = {}^0_{+1}\beta \)? Wait — that’s positron emission, but Bi decays to Pb via alpha or beta?

Wait: Bi → Pb — atomic number decreases from 83 to 82 → loss of one proton.

But in this case, mass stays same → likely emission of a positron or electron capture.

But let's check:
\( {}^{205}_{83}\text{Bi} \rightarrow {}^{205}_{82}\text{Pb} + X \)

So, for charge: 83 → 82 + Z ⇒ Z = +1
For mass: 205 → 205 + A ⇒ A = 0

So \( X = {}^0_{+1}\beta \) → positron (\( \beta^+ \))

✔ Answer: \( {}^0_{+1}\beta \) (or \( \beta^+ \))

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c) \( {}^{26}_{12}\text{Mg} + {}^1_1\text{p} \rightarrow {}^4_2\alpha + X \)

Left side:
- Mass: 26 + 1 = 27
- Atomic number: 12 + 1 = 13

Right side:
- Alpha: 4 + A ⇒ A = 23
- Charge: 2 + Z ⇒ Z = 11

So \( X = {}^{23}_{11}\text{Na} \)

✔ Answer: \( {}^{23}_{11}\text{Na} \)

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d) \( {}^{238}_{92}\text{U} + {}^{12}_6\text{C} \rightarrow X + 6{}^1_0\text{n} \)

Left:
- Mass: 238 + 12 = 250
- Charge: 92 + 6 = 98

Right:
- 6 neutrons: 6×1 = 6 mass, 0 charge
- So X must have:
- Mass: 250 − 6 = 244
- Charge: 98 − 0 = 98

So \( X = {}^{244}_{98}\text{Cm} \) (Curium)

✔ Answer: \( {}^{244}_{98}\text{Cm} \)

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e) \( {}^{59}_{27}\text{Co} + {}^2_1\text{H} \rightarrow {}^{60}_{27}\text{Co} + X \)

Left:
- Mass: 59 + 2 = 61
- Charge: 27 + 1 = 28

Right:
- Co-60: mass 60, charge 27
- So X must have:
- Mass: 61 − 60 = 1
- Charge: 28 − 27 = 1

So \( X = {}^1_1\text{H} \) → proton

✔ Answer: \( {}^1_1\text{H} \)

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f) \( {}^{97}_{43}\text{Tc} \rightarrow {}^{97}_{42}\text{Mo} + X \)

Tc → Mo: atomic number drops from 43 to 42 → loss of 1 proton

Mass: 97 → 97 + A ⇒ A = 0
Charge: 43 → 42 + Z ⇒ Z = +1

So \( X = {}^0_{+1}\beta \) → positron

✔ Answer: \( {}^0_{+1}\beta \)

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g) \( {}^{53}_{24}\text{Cr} + {}^4_2\alpha \rightarrow {}^1_0\text{n} + X \)

Left:
- Mass: 53 + 4 = 57
- Charge: 24 + 2 = 26

Right:
- Neutron: 1 mass, 0 charge
- So X: mass = 57 − 1 = 56, charge = 26 − 0 = 26

So \( X = {}^{56}_{26}\text{Fe} \)

✔ Answer: \( {}^{56}_{26}\text{Fe} \)

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h) \( {}^{239}_{94}\text{Pu} \rightarrow {}^{235}_{92}\text{U} + X \)

Mass: 239 → 235 + A ⇒ A = 4
Charge: 94 → 92 + Z ⇒ Z = 2

So \( X = {}^4_2\alpha \) → alpha particle

✔ Answer: \( {}^4_2\alpha \)

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i) \( {}^{20}_8\text{O} \rightarrow {}^{20}_9\text{F} + X \)

Mass: 20 → 20 + A ⇒ A = 0
Charge: 8 → 9 + Z ⇒ Z = −1

So \( X = {}^0_{-1}\beta \) → beta minus

✔ Answer: \( {}^0_{-1}\beta \)

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j) \( X + {}^4_2\text{He} \rightarrow {}^{17}_8\text{O} + {}^1_1\text{H} \)

Right side:
- Mass: 17 + 1 = 18
- Charge: 8 + 1 = 9

Left:
- He: 4, 2
- So X must have:
- Mass: 18 − 4 = 14
- Charge: 9 − 2 = 7

So \( X = {}^{14}_7\text{N} \)

✔ Answer: \( {}^{14}_7\text{N} \)

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k) \( {}^{26}_{12}\text{Mg} + X \rightarrow {}^{24}_{11}\text{Na} + {}^4_2\text{He} \)

Right:
- Mass: 24 + 4 = 28
- Charge: 11 + 2 = 13

Left:
- Mg: 26, 12
- So X must have:
- Mass: 28 − 26 = 2
- Charge: 13 − 12 = 1

So \( X = {}^2_1\text{H} \) → deuterium (heavy hydrogen)

✔ Answer: \( {}^2_1\text{H} \)

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l) \( {}^{252}_{98}\text{Cf} \rightarrow {}^{142}_{56}\text{Ba} + X + 4{}^1_0\text{n} \)

Left:
- Mass: 252
- Charge: 98

Right:
- Ba: 142, 56
- 4 neutrons: 4×1 = 4 mass, 0 charge
- So X must have:
- Mass: 252 − 142 − 4 = 106
- Charge: 98 − 56 − 0 = 42

So \( X = {}^{106}_{42}\text{Mo} \)

✔ Answer: \( {}^{106}_{42}\text{Mo} \)

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m) \( {}^{235}_{92}\text{U} \rightarrow X + 7{}^4_2\text{He} + 4{}^0_{-1}\beta \)

Right side:
- 7 He: 7×4 = 28 mass, 7×2 = 14 charge
- 4 β⁻: 0 mass, 4×(−1) = −4 charge
- So total right:
- Mass: A + 28 + 0 = A + 28
- Charge: Z + 14 − 4 = Z + 10

Left:
- U: 235 mass, 92 charge

So:
- Mass: 235 = A + 28 ⇒ A = 207
- Charge: 92 = Z + 10 ⇒ Z = 82

So \( X = {}^{207}_{82}\text{Pb} \)

✔ Answer: \( {}^{207}_{82}\text{Pb} \)

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✔ Final Answers Summary:

a) \( {}^{97}_{40}\text{Zr} \)
b) \( {}^0_{+1}\beta \)
c) \( {}^{23}_{11}\text{Na} \)
d) \( {}^{244}_{98}\text{Cm} \)
e) \( {}^1_1\text{H} \)
f) \( {}^0_{+1}\beta \)
g) \( {}^{56}_{26}\text{Fe} \)
h) \( {}^4_2\alpha \)
i) \( {}^0_{-1}\beta \)
j) \( {}^{14}_7\text{N} \)
k) \( {}^2_1\text{H} \)
l) \( {}^{106}_{42}\text{Mo} \)
m) \( {}^{207}_{82}\text{Pb} \)

Let me know if you'd like explanations in a table format!