Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Radioactive decay follows an exponential law where the quantity of a substance remaining after time $t$ is expressed as $N(t) = N_0 e^{-\lambda t}$. At time $t_1$, the decay of one-third leaves two-thirds of the original amount, yielding the relation $\frac{2}{3}N_0 = N_0 e^{-\lambda t_1}$, while at time $t_2$, the decay of two-thirds leaves one-third of the original amount, expressed as $\frac{1}{3}N_0 = N_0 e^{-\lambda t_2}$. Dividing these equations results in the ratio $\frac{1/3}{2/3} = \frac{e^{-\lambda t_2}}{e^{-\lambda t_1}}$, which simplifies to the expression $\frac{1}{2} = e^{-\lambda (t_2 - t_1)}$. Because this mathematical relationship indicates the substance has reduced to exactly half of its quantity at $t_1$, the duration $(t_2 - t_1)$ is equivalent to one half-life, which is 20 minutes.

This problem is governed by the principle of conservation of energy, where the total mass-energy of the parent nucleus converts into the total rest mass energy and kinetic energy of the resulting decay products. Initially, the system possesses an energy of $(M+\Delta m)c^2$. After the decay, the two daughter nuclei each have a mass of $\frac{M}{2}$, resulting in a total rest mass energy of $2 \times (\frac{M}{2})c^2 = Mc^2$. The mass difference $\Delta m$ is transformed into the total kinetic energy of the two nuclei, which is expressed as $2 \times (\frac{1}{2} \cdot \frac{M}{2} v^2) = \frac{M}{2}v^2$. By equating the initial total energy to the sum of the final rest mass energy and the kinetic energy, we obtain the balance $(M+\Delta m)c^2 = Mc^2 + \frac{M}{2}v^2$. Subtracting $Mc^2$ from both sides simplifies the expression to $\Delta mc^2 = \frac{M}{2}v^2$. Solving this for $v$ leads to $v^2 = \frac{2\Delta mc^2}{M}$, which gives $v = c\sqrt{\frac{2\Delta m}{M}}$.

The stability of an atomic nucleus is intrinsically linked to its binding energy per nucleon, with higher values representing a more stable energy state. When a parent nucleus decays into smaller fragments, the process is energetically favorable because the system shifts toward a configuration of greater stability, which necessitates that the binding energy per nucleon of the daughter nuclei, $E_2$, must exceed that of the parent nucleus, $E_1$.

Radioactive decay processes follow conservation laws where the emission of 3 alpha particles reduces the mass number by 12 and the atomic number by 6, while the subsequent emission of 2 positrons further reduces the atomic number by 2 without changing the mass. Consequently, the final nucleus possesses a mass number of $A-12$ and an atomic number of $Z-8$, which corresponds to the final count of protons. Calculating the number of neutrons by subtracting the proton count from the mass number yields $A - 12 - (Z - 8)$, which simplifies to $A - Z - 4$. This results in a final ratio of $\frac{A - Z - 4}{Z - 8}$ between neutrons and protons.

In any nuclear reaction, energy is released if the binding energy per nucleon of the product nuclei is greater than that of the reactant nuclei, signifying that the system has transitioned to a more stable state. For the fusion process $A + B \rightarrow C + \varepsilon$, the energy release $\varepsilon$ is positive because the binding energy per nucleon of $C$ is greater than that of $A$ and $B$. Similarly, for the fission reaction $F \rightarrow D + E + \varepsilon$, the value of $\varepsilon$ is positive because the binding energy per nucleon of the resulting nuclei $D$ and $E$ is greater than that of the reactant $F$. These two processes illustrate the physical conditions required for a net positive energy output in nuclear transformations.

The energy released during an electronic transition is proportional to the energy gap between the specific quantum states, with larger gaps producing higher-energy ultraviolet radiation and smaller gaps yielding lower-energy infrared radiation. Given that the transition from $n=4$ to $n=3$ results in ultraviolet light, any transition resulting in infrared radiation must involve a smaller energy change. Comparing the potential transitions shows that the jump from $n=5$ to $n=4$ possesses the smallest energy difference, characterizing it as the source of infrared radiation.

To determine the electron's properties, we equate the provided central force to the centripetal force required for circular motion, $\frac{k}{r} = \frac{m v^2}{r}$. This equality simplifies to $m v^2 = k$, which indicates that the kinetic energy $T = \frac{1}{2} m v^2 = \frac{k}{2}$ is entirely independent of the quantum number $n$. By incorporating Bohr's angular momentum quantization condition $m v r = \frac{n h}{2 \pi}$ and observing that the orbital velocity $v$ remains constant under this force, we find that the radius $r_n$ must vary directly with $n$.

Energy release in nuclear processes is driven by the system's shift toward greater stability, characterized by an increase in binding energy per nucleon. Since both the fission of heavy nuclei and the fusion of light nuclei result in products with higher binding energy per nucleon than the original reactants, Statement- 1 is correct. Conversely, Statement- 2 is false because the binding energy per nucleon actually decreases for heavy nuclei as atomic number $Z$ increases, whereas for light nuclei, it increases with increasing $Z$.

The rate of radioactive decay is determined by the product of the decay constant and the number of atoms present, expressed by the relationship $-\frac{dN}{dt} = \lambda N$. Given that the half-life of element $X$ is equal to the mean life of element $Y$, we set their mathematical definitions equal as $\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$, which simplifies to show that $\lambda_X = 0.693 \lambda_Y$. This indicates that the decay constant of element $Y$ is greater than that of element $X$. Since both elements begin with the same number of atoms, the higher decay constant of $Y$ results in a faster rate of decay compared to $X$.

Gamma ray emission is a mechanism where a nucleus transitions from an excited state to a lower energy state by releasing excess energy. Since these rays are forms of electromagnetic radiation lacking both electric charge and mass, they do not carry away any nucleons during the decay process. As a result, the proton number and the neutron number of the nucleus remain completely unchanged after the emission.

Photons are emitted only when an electron drops from a higher energy level to a lower one, which immediately restricts the possibilities to the transitions from $n=6$ to $n=2$ and from $n=2$ to $n=1$. The frequency of the emitted photon is directly proportional to the energy difference between these levels, governed by the relationship $\nu \propto (\frac{1}{n_{lower}^2} - \frac{1}{n_{higher}^2})$.

Calculating this proportionality constant for the transition from $n=6$ to $n=2$ gives $(\frac{1}{2^2} - \frac{1}{6^2}) = (\frac{1}{4} - \frac{1}{36}) = \frac{8}{36}$, which simplifies to $\frac{2}{9}$. In contrast, the transition from $n=2$ down to $n=1$ yields $(\frac{1}{1^2} - \frac{1}{2^2}) = (1 - \frac{1}{4}) = \frac{3}{4}$. Since $\frac{3}{4}$ is significantly greater than $\frac{2}{9}$, the transition between the two lowest energy levels results in the emission of a photon with the highest frequency.

The nuclear binding energy is defined as the energy equivalent of the mass defect, which represents the difference between the total mass of the individual constituent nucleons and the actual mass of the nucleus. For the oxygen isotope ${}_8\mathrm{O}^{17}$, the nucleus consists of 8 protons and 9 neutrons, so the mass defect is calculated by summing the masses of these 8 protons and 9 neutrons and subtracting the oxygen isotope mass $M_O$. Multiplying this mass difference by the square of the speed of light leads to the energy expression $(M_O - 8 M_P - 9 M_N) C^2$, which accounts for the mass of the isotope relative to its constituent components.

The distance of closest approach is reached when the alpha particle's initial kinetic energy is entirely converted into electrostatic potential energy as it is repelled by the target nucleus. By setting the kinetic energy $\frac{1}{2} m v^2$ equal to the potential energy $\frac{K(2e)(Ze)}{r}$, where $K$ is the Coulomb constant, we can solve for the distance $r$ to obtain the expression $r = \frac{4 K Z e^2}{m v^2}$. This relationship explicitly shows that the distance of closest approach is inversely proportional to the square of the velocity $v$, such that $r \propto \frac{1}{v^2}$.

To determine the emitted particle, we apply the laws of conservation of atomic number and mass number, ensuring these totals remain the same before and after the collision. Adding the components of the reactants, ${}_3^7\mathrm{Li} + {}_1^1\mathrm{p}$, results in a total atomic number of 4 and a total mass number of 8. Because these values already equate exactly to the ${}_4^8\mathrm{Be}$ nucleus produced, the additional emission must possess zero mass and zero charge, which identifies it as a gamma photon.

Radiation safety relies on quantifying the absorbed dose, which represents the amount of energy deposited by radiation into a specific mass of biological tissue. This dose is measured in rads, where the relationship is defined as $1 \text{ rad} = 100 \text{ ergs/g}$ of tissue. Since this specific absorption of energy is the primary factor that drives the potential harm and biological effect radiation has on the human body, the rad serves as the standard unit used to assess these consequences.

Nuclear reactions obey the principle of conservation of energy, where the initial kinetic energy of the proton added to the total binding energy of the reactants must equal the total binding energy of the products. The energy balance is defined by the formula $E + 7 \times 5.60 = 2 \times 4 \times 7.06$, where $E$ represents the kinetic energy of the proton. Simplifying the values results in the equation $E + 39.2 = 56.48$, which identifies the required proton energy as $E = 17.28$ MeV.

Unlike $\alpha$-decay, where particles are emitted with discrete energies, the energy released in $\beta$-decay is shared between an electron and an antineutrino, resulting in a continuous energy distribution. This process dictates that $\beta$-particles can be emitted with any kinetic energy ranging from zero up to a specific maximum value, $E_{max}$. Consequently, the spectrum $N(E)$ reflects this continuous range by starting at zero, increasing to a peak, and eventually returning to zero at the upper limit $E_{max}$.

Gamma radiation intensity follows the exponential decay law $I = I_0 e^{-\mu x}$, where the attenuation is determined by the number of half-value layers the radiation passes through. Since the initial intensity $I$ is reduced to $I/8$, and recognizing that $I/8 = I(1/2)^3$, it is evident that the radiation has traversed three half-value layers. Given that 36 mm of lead accounts for these three layers, a single half-value layer, which is the thickness required to reduce the intensity to $I/2$, is 36 mm divided by 3, which equals 12 mm.

Nuclear reactions obey the fundamental laws of conservation of mass number and atomic number, meaning the totals for both must balance on either side of the arrow. The shorthand notation $X(n,\alpha) \, {}^{7}_{3}\mathrm{Li}$ indicates that the unknown nucleus $X$ absorbs a neutron to produce a lithium nucleus and an alpha particle, which is represented by a helium nucleus as ${}^{4}_{2}\mathrm{He}$. We can write this transformation as the equation ${}^{A}_{Z}\mathrm{X} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{7}_{3}\mathrm{Li} + {}^{4}_{2}\mathrm{He}$. By balancing the mass numbers, we find $A + 1 = 7 + 4$, which leads to $A = 10$, and by balancing the atomic numbers, we calculate $Z + 0 = 3 + 2$, which gives $Z = 5$, identifying the nucleus as ${}^{10}_{5}\mathrm{B}$.

The energy emitted as a photon during an electron transition follows the relation $\Delta E = R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Because this energy difference is maximized when the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ is as large as possible, the jump from the $n=2$ level to the $n=1$ level yields the highest photon energy. Consequently, the transition marked as iii is the one that releases the most energetic photon among the choices.