NEET Nuclei Previous Year Questions with Solutions

Sum the masses of the thorium daughter nucleus and the alpha particle:

• • Carry all decimal places carefully: 234.043 + 4.003 = 238.046 u.

Subtract product mass from the parent uranium mass:

• • Positive mass defect confirms the alpha decay is energetically spontaneous (exothermic).

Multiply mass defect by energy equivalent per atomic mass unit:

• • Convert MeV to keV: 1 MeV = 1000 keV.
• • Q = 3.726 * 1000 keV = 3726 keV.
• • Matches Option (1) exactly.

An alpha particle is a helium-4 nucleus (⁴₂He). Emission of an alpha particle reduces the mass number by 4 and atomic number by 2:

In positron emission, a proton inside the nucleus transforms into a neutron, emitting a positron (⁰₊₁e) and a neutrino. The mass number remains unchanged, while the atomic number decreases by 1:

In beta-minus emission, a neutron transforms into a proton, emitting an electron (⁰₋₁e) and an antineutrino. The mass number is unchanged, while the atomic number increases by 1:

The final emission is an electron (e⁻, identical to β⁻). The mass number remains unchanged, and the atomic number increases by another 1:

The resulting product nucleus Q has mass number A = 286 and atomic number Z = 81.

The activity A of a radioactive sample at time t is related to its initial activity A₀ through the number of elapsed half-lives n:

Given that the activity drops to 1/16th of its initial value:

Multiply the number of half-lives by the duration of one half-life:

The parent nucleus has mass number $A = 240$ and binding energy per nucleon $\text{BE}_i/A = 7.6\text{ MeV}$:

The parent splits into two equal fragments of mass number $A = 120$ each, with $\text{BE}_f/A = 8.5\text{ MeV}$:

The total gain in binding energy (which is also the Q-value energy released in the fission reaction) is:

In all nuclear reactions, the total electric charge (atomic number $Z$) is strictly conserved. Let $\Delta Z$ be the change in atomic number of the daughter nucleus relative to the parent nucleus.

The atomic number changes from $Z$ to $Z-1$ (decreases by 1). This occurs when a proton converts into a neutron, emitting a positron ($\beta^+$):

The atomic number changes from $Z-1$ to $Z-3$ (decreases by 2). This occurs when an alpha particle (${}_2^4\text{He}$) is emitted:

The atomic number changes from $Z-3$ to $Z-2$ (increases by 1). This occurs when a neutron converts into a proton, emitting an electron ($\beta^-$):

The activity $A(t) = -\frac{dN}{dt} = \lambda N(t)$ follows the same exponential decay law as the number of undecayed nuclei. In terms of half-life $T_{1/2}$:

Given half-life $T_{1/2} = 100\text{ hours}$ and elapsed time $t = 150\text{ hours}$:

Substitute $n = 3/2$ into the decay equation:

Einstein showed that mass is another form of energy and that mass and energy can be converted into each other according to the famous relation: $$E = m c^2$$ where $$m$$ is mass in kilograms and $$c$$ is the speed of light in vacuum in meters per second.

Given mass in grams: $$m = 0.5\text{ g} = 0.5 \times 10^{-3}\text{ kg} = 5.0 \times 10^{-4}\text{ kg}$$

Using the standard value for the speed of light $$c = 3.0 \times 10^8\text{ m s}^{-1}$$: $$c^2 = (3.0 \times 10^8)^2 = 9.0 \times 10^{16}\text{ m}^2\text{ s}^{-2}$$ $$E = (0.5 \times 10^{-3}\text{ kg}) \times (9.0 \times 10^{16}\text{ m}^2\text{ s}^{-2})$$ $$E = 4.5 \times 10^{13}\text{ Joules (J)}$$

This is an enormous release of energy ($$4.5 \times 10^{13}\text{ J}$$, equivalent to about 10 kilotons of TNT) released from merely half a gram of annihilated matter.

When a uranium-235 nucleus absorbs a slow thermal neutron, it undergoes nuclear fission according to the reaction: $$^{235}_{92}\text{U} + ^1_0\text{n} \to ^{89}_{36}\text{Kr} + ^A_Z\text{X} + 3 ^1_0\text{n}$$ where $$^A_Z\text{X}$$ is the unknown co-product nucleus.

Total charge on reactant side equals total charge on product side: $$\sum Z_{\text{reactants}} = \sum Z_{\text{products}}$$ $$92 + 0 = 36 + Z + 3(0)$$ $$92 = 36 + Z$$ $$Z = 92 - 36 = 56$$ The element with atomic number $$Z = 56$$ is Barium ($$\text{Ba}$$).

Total mass number on reactant side equals total mass number on product side: $$\sum A_{\text{reactants}} = \sum A_{\text{products}}$$ $$235 + 1 = 89 + A + 3(1)$$ $$236 = 89 + A + 3$$ $$236 = 92 + A$$ $$A = 236 - 92 = 144$$

The missing nuclide is barium-144, represented as $$^{144}_{56}\text{Ba}$$. This matches Option (1).