Q.When ₃Li⁷ is bombarded by protons and the resultant nuclei are ₄Be⁸, the emitted particles are :

Mass number and atomic number already balance between reactants and product, so only energy (a gamma photon) is released — no material particle.

The reaction is:

$_3^7\text{Li} + \ _1^1\text{p} \rightarrow \ _4^8\text{Be} + X$

Check conservation of mass number (A) and atomic number (Z):

• Mass number: $7 + 1 = 8$, and $_4^8\text{Be}$ already has mass number 8 — balanced.
• Atomic number: $3 + 1 = 4$, and $_4^8\text{Be}$ already has atomic number 4 — balanced. …

Conserve mass number and atomic number across the nuclear reaction $_4Be^9 + \,_2He^4 \rightarrow X + \,_0n^1$.

Mass number: $9+4 = 13 = A_X + 1 \Rightarrow A_X = 12$

Atomic number: $4+2 = 6 = Z_X + 0 \Rightarrow Z_X = 6$

…

Correct α-decay: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He.

In α-decay a nucleus emits a ⁴₂He nucleus, so the mass number decreases by 4 and the atomic number by 2. Applying conservation:

• Mass number: 238 = 234 + 4 ✓
• Atomic number: 92 = 90 + 2 ✓ …

Both reactions are radioactive beta decays. Reaction (a) is $\beta^-$ decay (a neutron converts to a proton, emitting an electron and an antineutrino); reaction (b) is $\beta^+$ decay (a proton converts to a neutron, emitting a positron and a neutrino).

(a) ${}^{32}_{15}\text{P}\rightarrow{}^{32}_{16}\text{S}+\_\_\_\_+\bar\nu$

Mass number stays $32\to32$ (unchanged), but atomic number increases $15\to16$. An increase in $Z$ by 1 with no change in $A$, accompanied by an antineutrino, is the signature of $\beta^-$ (electron) decay:

$n \rightarrow p + e^- + \bar\nu$

So:

${}^{32}_{15}\text{P} \rightarrow {}^{32}_{16}\text{S} + {}^{0}_{-1}e + \bar\nu$

(b) ${}^{22}_{11}\text{Na}\rightarrow{}^{22}_{10}\text{Ne}+\_\_\_\_+\nu$

…

$Z = 2$ gives 2 protons; $A - Z = 4 - 2 = 2$ gives 2 neutrons: option (B).

Nucleons are the particles inside a nucleus — protons and neutrons (electrons are NOT nucleons).

For helium $_2He^4$:

• Atomic number $Z = 2$ → number of protons $= 2$. …

This is a deuterium-deuterium fusion reaction; balancing mass and atomic numbers shows a free neutron is released.

Check conservation of mass number (A) and atomic number (Z):

Left side: ${}_1H^2 + {}_1H^2$ → $A=2+2=4$, $Z=1+1=2$

Right side: ${}_2He^3$ → $A=3$, $Z=2$ …

Balancing mass number and atomic number across the nuclear reaction shows the emitted particle must be a proton.

The reaction is:

7N14 + 0n1 -> 6C14 + X

Conserve mass number (A): 14 + 1 = 14 + A_X, so A_X = 1.

Conserve atomic number (Z): 7 + 0 = 6 + Z_X, so Z_X = 1.

…

Conserving mass number and charge, W has A = 1 and Z = 0 → a neutron: option (c).

In a nuclear reaction, both the mass number (superscript) and the atomic number/charge (subscript) must be conserved.

Reaction: ₂He⁴ + ₓX^A → ₍Z₊₂₎Y^(A+3) + W.

Mass number balance:

4 + A = (A + 3) + (mass number of W)

⇒ mass number of W = 4 + A − A − 3 = 1.

Charge (atomic number) balance:

2 + Z = (Z + 2) + (charge of W) …

Neutrons = mass number − atomic number = A − Z.

The atomic number Z is the number of protons. The mass number A counts protons plus neutrons (the nucleons):

…

Nuclear reactions conserve charge, linear momentum and angular momentum (and energy, nucleon number).

The fundamental conservation laws hold in every nuclear reaction:

• Total electric charge is conserved.
• Total linear momentum is conserved.
• Total angular momentum is conserved. …

A controlled chain reaction requires the multiplication factor K = 1.

The neutron multiplication factor K is the ratio of the number of neutrons (or fissions) in one generation to that in the preceding generation:

$K = \frac{\text{neutrons produced in a generation}}{\text{neutrons in the preceding generation}}$

• If $K > 1$, the number of fissions grows each generation → the reaction is uncontrolled (as in a bomb).
• If $K < 1$, the reaction dies out. …