Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

The mass number of a nucleus represents the total sum of protons and neutrons contained within it, while the atomic number counts only the number of protons. Since ordinary hydrogen nuclei consist of a single proton with zero neutrons, its mass number and atomic number are both equal to one. For all other heavier nuclei, the presence of one or more neutrons ensures that the mass number is strictly greater than the atomic number.

Beta-minus decay is a radioactive transformation in which an unstable nucleus converts a neutron into a proton, simultaneously emitting an energetic electron and an electron antineutrino. In the given reaction, the parent nucleus $X$ transforms into a daughter nucleus $Y$ whose atomic number increases by one while its mass number remains unchanged. The simultaneous appearance of an electron and an antineutrino among the products uniquely characterizes this process as beta decay.

Radioactive decay follows an exponential reduction in mass over time, where the remaining mass after a given number of half-lives is expressed by the formula $M = M_0 \left(\frac{1}{2}\right)^n$. Starting with an initial mass of 10 g at time zero, the sample continuously decays as time progresses. After one half-life, the mass is reduced to 5 g, and after a second half-life, it halves once more to reach 2.50 g.

The spatial scale of an atom is defined by the radius of its electron cloud, which is typically on the order of $10^{-10} \,\text{m}$, whereas an atomic nucleus has a much smaller radius on the order of $10^{-15} \,\text{m}$. Since the volume of a sphere scales with the cube of its radius, the ratio of their volumes is found by cubing the ratio of their linear dimensions. Evaluating $\left(\frac{10^{-10}}{10^{-15}}\right)^3$ results in a volume factor of $10^{15}$, illustrating that an atom consists overwhelmingly of empty space.

Nuclear reactions must strictly conserve both total mass number and total atomic number across reactants and products.

$_8\text{O}^{16} + _1\text{H}^2 \rightarrow _7\text{N}^{14} + _2\text{He}^4$

When oxygen-16 is bombarded with a deuteron, the combined intermediate system temporarily possesses a mass number of 18 and an atomic number of 9. The emission of an alpha particle removes a mass number of 4 and an atomic number of 2, leaving behind a product nucleus with a mass number of 14 and an atomic number of 7, which identifies nitrogen-14 as the resulting element.

Radioactive decay proceeds geometrically, where half of the remaining active nuclei disintegrate during each successive half-life period. With a half-life of 10 days, a total duration of 30 days corresponds exactly to three half-life intervals. During the first 10 days, half of the initial 4 x 10^16 nuclei decay, leaving 2 x 10^16 active nuclei and generating an equal number of decayed nuclei. In the subsequent 10 days, half of the remaining active nuclei decay to add another 1 x 10^16 decayed nuclei, and during the final 10 days, half of what remains decays again to add 0.5 x 10^16 decayed nuclei. Summing the decayed amounts across all three intervals yields a total of 3.5 x 10^16 decayed nuclei.

Nuclear fusion occurs when extremely light atomic nuclei combine to form a heavier, more stable nucleus, releasing a tremendous amount of energy in the process. This process is energetically favorable because light nuclei possess a lower binding energy per nucleon compared to elements in the middle of the periodic table, making them unstable and prone to joining together. As these light particles fuse, the resulting nucleus achieves a higher binding energy per nucleon and a significantly more stable configuration.

Alpha rays consist of high-speed streams of helium nuclei that inherently carry a net positive electrical charge due to containing two protons without orbiting electrons. In contrast, beta rays comprise negatively charged electrons or positrons, while gamma rays and X-rays are neutral forms of electromagnetic radiation. This specific particulate composition causes alpha rays to deflect distinctly in electric and magnetic fields in accordance with positive charge behavior.

X-ray diffraction in crystals is described by Bragg's law, expressed as $2d \sin\theta = n\lambda$, where $d$ is the interplanar distance, $\theta$ is the glancing angle, $n$ is the order of diffraction, and $\lambda$ is the wavelength. To obtain the maximum possible wavelength for diffraction, the sine function must take its highest possible value of one, and the lowest order of diffraction $n = 1$ is considered. Substituting these conditions into the equation simplifies it to $\lambda_{\max} = 2d$. Multiplying the given interplanar distance by two yields the maximum wavelength of $5.6 \times 10^{-8}$ meters.

Nuclear fission releases a massive amount of energy because a fraction of the total rest mass of the heavy nucleus is converted directly into energy during the splitting process, governed by Einstein's mass-energy equivalence relation. When an unstable heavy nucleus divides into lighter daughter fragments, the combined mass of the resulting products is slightly less than the mass of the original parent nucleus. This missing mass, known as the mass defect, is expelled in the form of kinetic energy and radiation.

The amount of a radioactive substance remaining after a given time is governed by the decay equation $N = N_0 \left(\frac{1}{2}\right)^n$, where $n$ represents the number of half-life periods elapsed. Substituting the initial and remaining quantities into this relation yields $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$, leading to $\frac{1}{256} = \left(\frac{1}{2}\right)^n$. Equating the powers shows that $n = 8$, meaning eight half-lives must pass. Multiplying this number by the duration of a single half-life gives the total elapsed time of 100 hours.

The total energy of an electron in a hydrogen-like atom or ion is proportional to the square of the atomic number and inversely proportional to the square of the principal quantum number, following the relation $E \propto \frac{Z^2}{n^2}$. For a regular hydrogen atom, the atomic number is one, whereas for a singly ionized helium atom, the atomic number is two. Since both atoms are considered in the same orbit, the quantum number remains constant while the atomic number squared contributes a factor of four. Consequently, the energy of the helium ion is four times greater than that of the hydrogen atom in the corresponding orbit.

The shorthand notation for nuclear reactions links the initial target nucleus, incident particle, emitted particle, and product nucleus through strict conservation laws for mass and charge numbers.

$_5^{10}\text{B} + _0^1\text{n} \rightarrow _2^4\text{He} + _3^7\text{Li}$

Balancing the known mass numbers and atomic numbers on both sides of the equation requires the unknown parent nucleus to provide a mass number of 10 and an atomic number of 5. This uniquely identifies the reactant symbol as boron-10.

The actual mass of a stable atomic nucleus is always slightly less than the arithmetic sum of the individual masses of its constituent protons and neutrons due to the mass defect converted into nuclear binding energy. When individual nucleons bind together to form a nucleus, a specific amount of mass is released as energy, establishing the strict inequality $M < N M_n + Z M_p$. Consequently, the total nuclear mass remains less than the combined mass of its separate independent nucleons.

Nuclear radius scales with the cube root of the mass number according to the relation $r = r_0 A^{1/3}$. Finding the ratio between the sulphur nucleus and the helium nucleus involves dividing their respective mass numbers raised to the one-third power. Substituting the mass number 32 for sulphur and 4 for helium yields the cube root of 32 over 4. Simplifying the fraction inside gives the cube root of 8, which evaluates to a factor of 2.

In $\beta^+$ decay, conservation laws dictate that the total mass number, atomic number, and lepton number must remain constant across the reaction. Applying conservation of mass number yields $11 = 11 + 0 + A_X$, which means the mass number of particle $X$ is zero. Similarly, conservation of atomic number gives $6 = 5 + 1 + Z_X$, indicating that particle $X$ carries zero charge. A subatomic particle with zero mass and zero charge that preserves lepton number during positron emission is precisely an electron neutrino.

The radius of any given orbit in the Bohr model scales proportionally with the square of the principal quantum number, expressed by the relation $r \propto n^2$. Substituting the principal quantum number for the second orbit into this proportionality gives $2^2$, which evaluates to $4$. Multiplying this scaling factor by the base Bohr radius yields an orbital radius of $4 a_{0}$.

The maximum energy of an X-ray photon produced by striking electrons equals the kinetic energy acquired by the electrons across an accelerating potential difference, expressed by the equation $\frac{hc}{\lambda} = eV$. Rearranging this fundamental relation to isolate the wavelength reveals that it depends inversely on the applied voltage. Since Planck's constant, the speed of light, and the elementary charge remain constant, the minimum wavelength scales directly with the reciprocal of the potential difference.

The centripetal force required for an electron's orbital motion is supplied entirely by the electrostatic attraction between the nucleus and the orbiting particle. Equating the mechanical centripetal force to Coulomb's electrostatic force yields the relation $\frac{m v^{2}}{a_{0}} = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{a_{0}^{2}}$. Rearranging this equation to solve for velocity isolates the variables, resulting \in $v = \frac{e}{\sqrt{4 \pi \varepsilon_{0} a_{0} m}}$.

Transitions ending at the principal quantum number $n = 2$ define the Balmer series in the atomic hydrogen spectrum. The first line of this series originates from an electron falling from $n = 3$ to $n = 2$, whereas the subsequent transition from $n = 4$ to $n = 2$ produces the second line. Therefore, an electron transition from $n = 4$ to $n = 2$ corresponds directly to the second line of the Balmer series.