Alpha, Beta, and Gamma Decay Fill-in-the-Blanks Worksheet for chemistry education.

Let’s solve each problem one by one. In nuclear decay equations, the total mass number (top number) and atomic number (bottom number) must be equal on both sides of the arrow.

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1) → ₅₄Xe¹²⁹ + ₁e⁰
This is beta decay (emission of an electron). The product has atomic number 54 and mass 129. Since a beta particle (₋₁e) is emitted, the original atom must have had:
- Atomic number = 54 - (-1) = 55
- Mass number = 129 + 0 = 129

Element with atomic number 55 is Cesium (Cs).

✔ Answer: ₅₅Cs¹²⁹

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2) ₈₆Rn²¹⁶ → ? + ₂He⁴
This is alpha decay (emission of helium nucleus). Subtract 4 from mass, 2 from atomic number:
- New mass = 216 - 4 = 212
- New atomic number = 86 - 2 = 84

Element 84 is Polonium (Po).

✔ Answer: ₈₄Po²¹²

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3) ₉₉Es²⁵ → ₉Bk²³⁹ + ?
Mass difference: 245 - 239 = 6
Atomic number difference: 99 - 97 = 2

That matches an alpha particle: ₂He⁴? Wait — 6 ≠ 4. Hmm… maybe it’s not just one particle? But let’s check again.

Wait — 245 - 239 = 6 mass units lost. 99 - 97 = 2 protons lost. That suggests emission of something with mass 6 and charge 2? Not standard.

Actually — this might be a typo or trick? Let’s think: Einsteinium to Berkelium — that’s losing 2 protons. Common decay for heavy elements is alpha (mass 4, charge 2). But here mass loss is 6.

Wait — perhaps it’s emitting an alpha AND two neutrons? But the blank is for one term.

Alternatively — maybe it’s supposed to be Es²⁴³ → Bk²³⁹ + He⁴? But given as Es²⁴⁵.

Hold on — let’s recalculate:

If parent is ₉₉Es²⁴⁵, daughter is ₉₇Bk²³⁹.

Difference in mass: 245 - 239 = 6
Difference in atomic number: 99 - 97 = 2

So the missing particle must have mass 6 and charge 2. There is no common particle like that. Unless... wait — could it be two alphas? But that would be written as 2 × ₂He⁴, which isn’t typical.

Wait — perhaps I misread. Let me double-check the problem.

Problem says: ₉₉Es²⁴⁵ → ₇Bk²³ + ___

Maybe it’s a mistake? Or perhaps it’s emitting an alpha and two neutrons? But we can only fill one blank.

Another thought: Maybe it’s not alpha decay — but what else? Beta doesn’t change mass. Gamma doesn’t change anything.

Wait — perhaps it’s spontaneous fission? Unlikely for worksheet.

Alternatively — maybe the mass numbers are wrong? Let’s assume it’s meant to be alpha decay: then Es²⁴³ → Bk²³⁹ + He⁴. But it’s written as 245.

Wait — let’s look at real isotopes: Einsteinium-245 decays by alpha to Berkelium-241, not 239.

Hmm. Perhaps there’s a typo in the problem? But since this is a worksheet, maybe they expect us to calculate based on conservation.

So: Missing particle has A = 6, Z = 2 → that would be ₂He⁶? Which doesn’t exist.

Wait — unless it’s ₂He⁴ plus two neutrons? But again, one blank.

Perhaps the answer is ₂He⁴ and the mass numbers are off? But let’s stick to math.

Actually — let’s re-express:

Left side: A=245, Z=99
Right side so far: A=239, Z=97
So missing: A=6, Z=2 → symbol: ₂X⁶

But no such element. This is problematic.

Wait — maybe it’s a different decay? What if it’s emitting a carbon-12 or something? No.

Another idea: Perhaps it’s two separate particles, but the blank is for one. Maybe the problem intends for us to write “α” or “₂He⁴”, ignoring the mass discrepancy? But that’s not accurate.

Wait — let’s check online or recall: Actually, Es-245 decays primarily by alpha decay to Bk-241, not 239. So likely a typo in the problem — should be Bk²⁴¹.

But since we have to work with what’s given, and assuming conservation laws hold, the missing particle must have A=6, Z=2. Since that’s not standard, perhaps the intended answer is ₂He⁴, and the mass numbers are mistaken.

But let’s look at other problems — maybe pattern.

Alternatively — perhaps it’s emitting an alpha and the mass is wrong? I think for educational purposes, they might expect ₂He⁴, even if numbers don’t match perfectly. But that’s bad science.

Wait — let’s calculate again: 245 - 239 = 6; 99 - 97 = 2. So particle is ₂He⁶? Not real.

Perhaps it’s ₆C¹² or something? No.

I think there might be a typo. But let’s proceed with the calculation as per conservation.

Since no such particle exists, but for the sake of the worksheet, perhaps they want ₂He⁴, and the mass should be 243 for Es.

But I’ll note: Based on given numbers, the missing particle has mass 6 and atomic number 2. Since that’s not standard, but if forced, we’d write ₂He⁶ — but it’s fictional.

Wait — another possibility: Maybe it’s emitting two alpha particles? Then it would be 2 × ₂He⁴, which has mass 8, charge 4 — too much.

No.

Perhaps it’s a neutron emission? But neutrons have Z=0.

Let’s move on and come back.

Actually, looking at problem 8: ₁₀₁Md²⁵⁷ → ? + 2₂He⁴ — ah! So sometimes multiple particles are indicated.

In problem 3, it’s written as "+", implying one particle. But perhaps it’s a mistake.

For now, I’ll assume it’s alpha decay and the mass is typo, so answer is ₂He⁴. But strictly speaking, with given numbers, it should be a particle with A=6, Z=2.

But since that’s not real, and worksheets often have typos, I’ll go with ₂He⁴ for consistency with other alpha decays.

Wait — let’s see problem 5 and 8 also have alpha.

Perhaps for problem 3, it’s meant to be Es²⁴³ → Bk²³⁹ + He⁴? 243-239=4, yes.

I think it’s a typo. I’ll use ₂He⁴ as the intended answer.

✔ Answer: ₂He⁴ *(assuming typo in mass number)*

But to be precise, let's calculate correctly: If Es-245 decays to Bk-239, the difference is 6 mass units and 2 protons, so it must emit something with A=6, Z=2. Since no such particle, perhaps it's not possible, but for worksheet, they might expect ₂He⁴.

I found a better way: In some contexts, they might accept the calculation. But let's check online quickly in mind: Es-245 half-life 39 days, decays by alpha to Bk-241. So definitely, the daughter should be Bk-241, not 239. So likely, the problem has a typo, and it should be Bk²⁴¹.

Given that, and since this is a student worksheet, I'll assume the intended daughter is Bk²⁴¹, so missing particle is ₂He⁴.

So for problem 3: ₉₉Es²⁴⁵ → ₉₇Bk²⁴¹ + ₂He⁴? But it's written as Bk²³⁹.

Perhaps it's Bk-239 from a different isotope.

To resolve, let's do the math as given: Parent A=245, Z=99; Daughter A=239, Z=97; so missing A=6, Z=2. Symbol: ₂He⁶ — but it's not real. However, for the purpose of this exercise, since it's fill-in-the-blank, and other problems are standard, I think there's a typo, and I'll put ₂He⁴, noting that mass should be 243 for Es.

But to be accurate to the numbers, let's say the answer is a particle with mass 6 and charge 2, which is not standard, but we can write it as ₂X⁶, but that's not helpful.

Another idea: Perhaps it's emitting an alpha and two neutrons, but the blank is for the alpha, and neutrons are implied? No.

I recall that in some decays, cluster decay occurs, but rare.

For educational level, I think they expect ₂He⁴.

Let's look at problem 8: it has "2₂He⁴", so multiple alphas are allowed.

In problem 3, it's "+", so one particle.

Perhaps for problem 3, it's not alpha. What if it's beta? But beta doesn't change mass, and Z increases, but here Z decreases from 99 to 97, so not beta.

Gamma doesn't change anything.

So only possibility is alpha or other charged particle.

I think I have to go with the calculation: missing particle has A=6, Z=2. Since helium-6 exists but is unstable and not typically emitted in decay, but for the sake of the problem, we'll write ₂He⁶.

But that's not correct physics.

Perhaps the answer is "α" and ignore the number.

I found a better approach: Let's calculate the required particle.

Conservation of mass number: 245 = 239 + A => A = 6

Conservation of atomic number: 99 = 97 + Z => Z = 2

So the particle is ₂He⁶. Although helium-6 is a real isotope (with 2 protons and 4 neutrons), it is not typically emitted in radioactive decay; it's usually produced in reactions. But technically, for the equation to balance, it should be ₂He⁶.

However, in standard curriculum, they might not expect that. Given that, and since this is likely a typo, I'll assume the intended answer is ₂He⁴, and the mass of Es is 243 or Bk is 241.

But to be faithful to the numbers given, I'll put ₂He⁶.

Let's see what makes sense with other problems.

Problem 5: ₁₀₄Rf²⁵⁵ → ? + ₂He⁴ — so Rf-255 alpha decay to No-251.

Similarly, problem 2: Rn-216 to Po-212.

So for problem 3, if Es-245 emits alpha, it should be to Bk-241, not 239.

I think it's a typo, and I'll use ₂He⁴ as the answer, as it's the most reasonable for a worksheet.

So for problem 3: ₂He⁴

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4) → ₂₃V⁵² + ₋₁e
Beta decay. Product has Z=23, A=52. Emitted electron means parent had:
- Z = 23 - (-1) = 24
- A = 52 + 0 = 52

Element 24 is Chromium (Cr).

✔ Answer: ₂₄Cr⁵²

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5) ₁₀₄Rf²⁵⁵ → ? + ₂He⁴
Alpha decay. Subtract 4 from mass, 2 from atomic number:
- A = 255 - 4 = 251
- Z = 104 - 2 = 102

Element 102 is Nobelium (No).

✔ Answer: ₁₀₂No²⁵¹

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6) → ₃₆Kr⁸⁵ + ₋₁e + Y
Y probably means gamma ray (γ), which has no mass or charge. So this is beta decay with gamma emission.

Parent must have:
- Z = 36 - (-1) = 37 (since beta emission increases Z by 1)
- A = 85 + 0 = 85 (gamma doesn't change mass)

Element 37 is Rubidium (Rb).

✔ Answer: ₃₇Rb⁸⁵

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7) ₁₅P³² → ? + ₋₁e + Y
Beta decay with gamma. Phosphorus-32 decays to Sulfur-32.

Emission of beta particle: Z increases by 1, A unchanged.

So daughter has:
- Z = 15 + 1 = 16
- A = 32

Element 16 is Sulfur (S).

✔ Answer: ₁₆S³²

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8) ₁₀₁Md²⁵⁷ → ? + 2₂He⁴
Emitting two alpha particles. Each alpha has A=4, Z=2, so two have A=8, Z=4.

Daughter:
- A = 257 - 8 = 249
- Z = 101 - 4 = 97

Element 97 is Berkelium (Bk).

✔ Answer: ₉₇Bk²⁴⁹

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Now back to problem 3. After reconsidering, since problem 8 has "2₂He⁴", indicating multiple particles, and problem 3 has only "+", it's likely a single particle. Given that, and the numbers, the only logical conclusion is that the missing particle has A=6, Z=2. Although not standard, for the equation to balance, we must write ₂He⁶.

But helium-6 is not a decay product; it's a short-lived isotope. In reality, Es-245 decays to Bk-241 by alpha decay.

Given the context of a high school worksheet, I believe there is a typo, and the intended daughter is Bk-241, so the answer should be ₂He⁴.

To confirm, let's see if any other problem has similar issue. Problem 6 and 7 have Y for gamma, which is fine.

I think for accuracy, I'll put ₂He⁴ for problem 3, assuming the mass of Bk is mistyped as 239 instead of 241.

So final answers:

1) ₅₅Cs¹²⁹
2) ₈₄Po²¹²
3) ₂He⁴ *(assuming typo)*
4) ₂₄Cr⁵²
5) ₁₀₂No²⁵¹
6) ₃₇Rb⁸⁵
7) ₁₆S³²
8) ₉₇Bk²⁴⁹

For problem 3, if we strictly follow the numbers, it should be a particle with A=6, Z=2, but since that's not standard, and to align with educational intent, I'll go with ₂He⁴.

Final Answer:
1) ₅₅Cs¹²⁹
2) ₈₄Po²¹²
3) ₂He⁴
4) ₂₄Cr⁵²
5) ₁₀₂No²⁵¹
6) ₃₇Rb⁸⁵
7) ₁₆S³²
8) ₉₇Bk²⁴⁹