Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

According to Einstein's mass-energy equivalence relation $E = mc^2$, mass can be converted into energy by multiplying by the square of the speed of light. When applying this fundamental conversion to one atomic mass unit, which represents one-twelfth of the mass of a carbon-12 atom, the resulting energy is expressed in electron-volts or mega-electron-volts. Specifically, one atomic mass unit is universally equivalent to $931 \ \mathrm{MeV}$ of energy.

Radioactive transformations systematically modify a nuclide's structural parameters based on the conservation laws governing particle emission. Emitting a single alpha particle from the parent nucleus $_{n}\mathrm{X}^{m}$ reduces its atomic number by two and its mass number by four, yielding $_{n-2}\mathrm{Y}^{m-4}$. Subsequent emission of two beta particles each increases the atomic number by one, which perfectly neutralizes the initial reduction of two. Thus, the resulting nucleus retains the original atomic number $n$ while its mass number decreases by four, producing $_{n}\mathrm{X}^{m-4}$.

Nuclides are classified based on their internal subatomic particle counts, specifically distinguishing between protons, total nucleons, and neutrons. Calculating the difference between the mass number and the atomic number for carbon-13 yields six protons and seven neutrons. Performing the exact same arithmetic for nitrogen-14 reveals seven protons and seven neutrons. Because both distinct atomic species share an identical neutron count despite having different atomic numbers, they belong to the category of isotones.

Nuclear stability is achieved through a phenomenon known as the mass defect, where a portion of the constituent nucleons' mass is converted into binding energy when they combine to form a nucleus. An alpha particle consists of two protons and two neutrons bound tightly together in a helium nucleus. Because energy is released during this binding process, the actual measured mass of the alpha particle is always slightly less than the sum of the individual masses of two isolated protons and two neutrons.

The radius of an electron's orbit in a hydrogen atom scales directly with the square of the principal quantum number according to the relation $r = n^2 r_0$, where $r_0$ represents the Bohr radius. When the atom is promoted to its first excited level, the principal quantum number becomes $n = 2$. Squaring this value results in an orbital radius that is four times larger than the ground state radius.

Characteristic X-rays are produced when high-energy electrons eject inner-shell electrons from the target atoms, leaving vacancies that are immediately filled by electrons dropping from outer shells. These precise electronic transitions release energy in the form of photons at specific, well-defined wavelengths, which appear as sharp peaks superimposed on the continuous X-ray background. Because the energies and wavelengths of these emissions depend exclusively on the atomic number of the target material, they uniquely identify the element and form the characteristic spectrum.

An alpha particle consists of two tightly bound protons and two neutrons, making it identical to the bare nucleus of a helium atom. Because it completely lacks orbital electrons, it carries a net positive charge and is formally classified as a doubly ionized helium atom. This specific subatomic structure distinguishes alpha particles from standalone protons, positrons, or neutral gas atoms.

Nuclear binding energy per nucleon represents the average energy required to dismantle a nucleus into its constituent protons and neutrons, reflecting overall nuclear stability. Across most stable medium-sized nuclei, the strong nuclear force binds nucleons tightly together, resulting in a stabilization energy that hovers around 8 MeV per nucleon.

The average atomic mass of an element is determined by the weighted average of the isotopic masses relative to their natural abundances. Letting the percentage of ${}_{5}\text{B}^{10}$ be $x$ and that of ${}_{5}\text{B}^{11}$ be $100 - x$, the average atomic mass equation is set up as $\frac{10x + 11(100 - x)}{100} = 10.81$. Solving this relation gives $x = 19$, which means the abundance of ${}_{5}\text{B}^{10}$ is $19\%$ and ${}_{5}\text{B}^{11}$ is $81\%$, producing the ratio $19:81$.

The energy required to ionize a single-electron atomic system scales quadratically with the nuclear charge, following the proportionality $E \propto Z^2$. Since a singly ionized helium atom has a nuclear charge of two, its ionization energy is four times greater than that of a standard hydrogen atom. Multiplying the baseline hydrogen ionization energy of 13.6 eV by this factor gives 54.4 eV.

Radioactive decay follows an exponential reduction pattern where the remaining count rate is expressed by the formula $N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$. The total elapsed time of 2 hours equals 120 minutes, which represents four half-life periods when divided by the 30-minute half-life. Substituting this into the decay fraction yields one-half to the power of four, or $\frac{1}{16}$, indicating the final rate is one-sixteenth of the original. Multiplying the final count rate of 5 by 16 gives the starting value of 80.

The energy of an electron in a given orbit of a hydrogen atom is inversely proportional to the square of the principal quantum number, expressed by the formula $E_n = \frac{-13.6}{n^2}$. Substituting the quantum number $n = 2$ into this expression yields $E_2 = \frac{-13.6}{2^2}$. Evaluating the denominator gives $4$, and dividing $-13.6$ by $4$ results in $-3.4$ eV.

The atomic nucleus is composed entirely of nucleons, which consist of positively charged protons and electrically neutral neutrons. Electrons reside outside the nucleus in discrete orbital shells and do not form any part of the nuclear structure. Therefore, only neutrons and protons make up the constituents of atomic nuclei.

In a hydrogen atom, potential energy is negative and inversely proportional to the orbital radius, expressed as $PE = -\frac{kZe^2}{r}$, while kinetic energy is positive and given by $KE = \frac{kZe^2}{2r}$. When an atom transitions to a higher excited state, its orbital radius increases, causing the absolute value of the negative potential energy to shrink, which actually means the potential energy itself increases. Simultaneously, the larger radius diminishes the kinetic energy since it scales inversely with distance.

The electrostatic attractive force between the nucleus and the revolving electron supplies the necessary centripetal force for circular motion. Equating the Coulomb force to the centripetal force gives the radius relation $r = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{mv^2}$. Substituting the specified values where the velocity squared is $25.6 \times 10^{10}$ and evaluating the fraction results in a radius of $1 \times 10^{-9} \text{ m}$, establishing the value as $1$.

The total number of possible spectral lines emitted when an electron cascades down from an excited state with principal quantum number $n$ is given by the formula $\frac{n(n-1)}{2}$. Substituting the principal quantum number $4$ into this expression accounts for every unique downward transition path between energy levels. Multiplying $4$ by $3$ gives $12$, which is then divided by $2$ to result \in a total of $6$ distinct spectral lines.

Radioactive decay sequences alter atomic and mass numbers according to the specific type of particle emitted. A single alpha emission reduces the mass number by four units and decreases the atomic number by two units. In contrast, each beta emission increases the atomic number by one unit while leaving the mass number unchanged, meaning two beta emissions provide a net increase of two to the atomic number. Consequently, the combined effect of one alpha and two beta emissions cancels out the change in atomic number entirely while leaving a net reduction in mass number.

Nuclear radius is directly proportional to the cube root of the mass number, which implies that the volume of a nucleus scales linearly with the mass number rather than $A^{1/3}$. Consequently, volume is proportional to $A$ itself. Additionally, mass defect is defined strictly as the mass difference between a bound nucleus and the combined mass of its individual constituent protons and neutrons, excluding any atomic electrons. Therefore, the statements affirming linear volume scaling with mass number and defining mass defect relative to nuclear constituents are accurate.

The energy levels of a hydrogen atom follow the principle that $E_n = \frac{E_0}{n^2}$, where the second excited state corresponds to the principal quantum number $n = 3$. Ionizing an atom from a specific excited state requires supplying an amount of energy equal in magnitude to the binding energy of that exact state. Dividing the ground state energy of 13.6 eV by the square of 3 yields the required ionization energy of 1.51 eV.

The energy released in a nuclear reaction, known as the $Q$-value, is determined by calculating the mass defect between the reactants and products and converting it into energy using mass-energy equivalence. Subtracting the combined mass of the product nuclei from the mass of the parent uranium nucleus yields a mass defect of $0.004 \text{ u}$. Multiplying this mass defect by the energy equivalent of an atomic mass unit gives $3.726 \text{ MeV}$, which converts to $3726 \text{ keV}$.