Worksheet for practicing nuclear equations involving alpha and beta decay.

You're given two sets of nuclear decay equations — alpha decay (Problem 1) and beta decay (Problem 2). We need to find the unknown element X in each case by conserving mass number (top number) and atomic number/proton number (bottom number).

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## 🔹 Problem 1: Alpha Decay

In alpha decay, the nucleus emits an alpha particle:
→ ⁴₂He (which has mass number 4, atomic number 2).

So for any decay:
Parent → Alpha + Daughter

We conserve:
- Mass number: total on left = total on right
- Atomic number: total on left = total on right

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a) ²²²₈₈Ra → ⁴₂He + ___X

- Mass: 222 = 4 + A → A = 218
- Atomic number: 88 = 2 + Z → Z = 86

Element with atomic number 86 is Radon (Rn).

✔ Answer: ²¹⁸₈₆Rn

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b) ²⁰⁸₉₂U → ⁴₂He + ___X

- Mass: 208 = 4 + A → A = 204
- Atomic number: 92 = 2 + Z → Z = 90

Element with atomic number 90 is Thorium (Th).

✔ Answer: ²⁰⁴₉₀Th

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c) ²⁰⁸₈₄Po → ⁴₂He + ___X

- Mass: 208 = 4 + A → A = 204
- Atomic number: 84 = 2 + Z → Z = 82

Element with atomic number 82 is Lead (Pb).

✔ Answer: ²⁰⁴₈₂Pb

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d) ¹⁸⁵₇₉Au → ⁴₂He + ___X

- Mass: 185 = 4 + A → A = 181
- Atomic number: 79 = 2 + Z → Z = 77

Element with atomic number 77 is Iridium (Ir).

✔ Answer: ¹⁸¹₇₇Ir

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e) ¹⁸⁵₇₅Pt → ⁴₂He + ___X

- Mass: 185 = 4 + A → A = 181
- Atomic number: 75 = 2 + Z → Z = 73

Element with atomic number 73 is Tantalum (Ta).

✔ Answer: ¹⁸¹₇₃Ta

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## 🔹 Problem 2: Beta Decay

In beta minus decay (β⁻), a neutron turns into a proton, emitting an electron:
→ ⁰₋₁e (mass number 0, atomic number -1)

So for any decay:
Parent → Electron + Daughter

Conservation rules:
- Mass number: unchanged (electron has mass ~0)
- Atomic number: increases by 1 (because neutron → proton adds +1 to atomic number)

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a) ⁶₂He → ⁰₋₁e + ___X

- Mass: 6 = 0 + A → A = 6
- Atomic number: 2 = -1 + Z → Z = 3

Element with atomic number 3 is Lithium (Li).

✔ Answer: ⁶₃Li

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b) ²⁴₁₁Na → ⁰₋₁e + ___X

- Mass: 24 = 0 + A → A = 24
- Atomic number: 11 = -1 + Z → Z = 12

Element with atomic number 12 is Magnesium (Mg).

✔ Answer: ²⁴₁₂Mg

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c) ²⁰¹₇₀Au → ⁰₋₁e + ___X

Wait — Gold (Au) has atomic number 79, not 70. This looks like a typo. But we’ll solve based on what’s written.

Assuming it's meant to be ²⁰¹₇₀? — element with Z=70 is Ytterbium (Yb).

But since the symbol says Au (gold), which is Z=79, there’s inconsistency. However, we follow the numbers given.

So:

- Mass: 201 = 0 + A → A = 201
- Atomic number: 70 = -1 + Z → Z = 71

Element with atomic number 71 is Lutetium (Lu).

✔ Answer: ²⁰¹₇₁Lu

*(Note: If this was meant to be gold, it should be ²⁰¹₇₉Au → ⁰₋₁e + ²⁰¹₈₀Hg)*

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d) ⁴²₁₉K → ⁰₋₁e + ___X

- Mass: 42 = 0 + A → A = 42
- Atomic number: 19 = -1 + Z → Z = 20

Element with atomic number 20 is Calcium (Ca).

✔ Answer: ⁴²₂₀Ca

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e) ¹⁴⁵₅₅Cs → ⁰₋₁e + ___X

- Mass: 145 = 0 + A → A = 145
- Atomic number: 55 = -1 + Z → Z = 56

Element with atomic number 56 is Barium (Ba).

✔ Answer: ¹⁴⁵₅₆Ba

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## ✔ Final Answers:

Problem 1 (Alpha Decay):

a) ²¹⁸₈₆Rn
b) ²⁰⁴₉₀Th
c) ²⁰⁴₈₂Pb
d) ¹⁸¹₇₇Ir
e) ¹⁸¹₇₃Ta

Problem 2 (Beta Decay):

a) ⁶₃Li
b) ²⁴₁₂Mg
c) ²⁰¹₇₁Lu *(assuming input is correct; if Au was intended, it should be ²⁰¹₇₉Au → ²⁰¹₈₀Hg)*
d) ⁴²₂₀Ca
e) ¹⁴⁵₅₆Ba

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Let me know if you’d like the reasoning explained differently or want to check for typos in the original problem!