NEET Nucleus MCQs & PYQs

In alpha decay ($A \rightarrow B$), the atomic number Z decreases by 2. In two beta decays ($B \rightarrow C$), the atomic number increases by $2 \times 1 = 2$. Thus, the final atomic number of C equals the initial atomic number of A. Elements with the same Z are isotopes.

Curie (Ci) is a non-SI unit of radioactivity, originally defined as the activity of 1 gram of radium-226 ($3.7 \times 10^{10}$ decays per second).

The number of half-lives $n = \frac{6400}{800} = 8$. The fraction remaining is $(1/2)^n = (1/2)^8 = 1/256$.

Going backward in time by two months corresponds to adding 2 half-lives to the activity. The activity was $2 \times 2^2 = 2 \times 4 = 8$ micro curies.

Gamma radiation involves the emission of high-energy photons when a nucleus transitions from a higher energy state to a lower one. This process does not alter the number of protons or neutrons, so the mass number and atomic number remain unchanged.

An $\alpha$-particle is identical to a helium-4 nucleus, which contains exactly 2 protons and 2 neutrons, with no electrons.

Emission of an $\alpha$ particle reduces mass number by 4 and atomic number by 2 (result: $^{m-4}_{n-2}X'$). Emission of two $\beta^-$ particles increases atomic number by 2 while leaving mass number unchanged (result: $^{m-4}_{n-2+2}Z = ^{m-4}_{n}Z$).

For every $x$ alpha particles, the atomic number decreases by $2x$. For $2x$ beta particles, the atomic number increases by $2x$. The net change in atomic number Z is zero. Nuclei with the same atomic number are isotopes.

Step 1: Z increases by 1, A is constant -> $\beta$ emission. Step 2: Z decreases by 2, A decreases by 4 -> $\alpha$ emission. Step 3: Excited state ($B^*$) to ground state ($B$) with no change in Z or A -> $\gamma$ emission. The sequence is $\beta, \alpha, \gamma$.

The fraction remaining is $25/100 = 1/4 = (1/2)^2$. This means 2 half-lives have passed. The time elapsed is $2 \times 1600 = 3200$ years.