Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Nuclear fusion releases energy because the product nucleus is more stable than the reactants, which is determined by calculating the difference in total binding energy. Two deuterium nuclei, each with two nucleons at a binding energy of 1.1 MeV per nucleon, result in a reactant total of 4.4 MeV. The single helium-4 product has four nucleons at 7.2 MeV each, totaling 28.8 MeV, meaning the energy released is:

$Q = 28.8 \text{ MeV} - 4.4 \text{ MeV} = 24.4 \text{ MeV}$

The distance of closest approach occurs when the initial kinetic energy ($K$) of the alpha particle is entirely converted into electrostatic potential energy, defined by the relationship $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. With the given parameters where $e = 1.6 \times 10^{-19}$ C, $Z = 79$, $K = 7.7 \times 10^6 \times 1.6 \times 10^{-19}$ J, and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ Nm²/C², the equation becomes:

$r_0 = \frac{2 \times (9 \times 10^9) \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}$

Simplifying the expression by canceling one term of $1.6 \times 10^{-19}$ and calculating the remaining factors:

$r_0 = \frac{2 \times 9 \times 10^9 \times 79 \times 1.6 \times 10^{-19}}{7.7 \times 10^6} \approx 2.95 \times 10^{-14} \text{ m}$

The total energy released in nuclear fission is determined by multiplying the number of atoms undergoing fission by the average energy released per fission event. The number of atoms, $N$, present in 47 g of $^{235}\mathrm{U}$ is calculated using the molar mass of 235 g/mol and Avogadro's number:

$N = \frac{47}{235} \times 6 \times 10^{23} = 1.2 \times 10^{23}$

With the total number of atoms established, the total energy released is calculated by multiplying $N$ by the average energy released per fission, which is 190 MeV:

$E = 1.2 \times 10^{23} \times 190 = 228 \times 10^{23} \text{ MeV}$

Comparing this calculated result with the given expression $\alpha \times 10^{23} \text{ MeV}$, it is clear that the value of $\alpha$ is 228.

The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant. Expressing the radii for the two nuclei as $R_{\alpha} = R_0 \alpha^{1/3}$ and $R_{\beta} = R_0 \beta^{1/3}$, and substituting $\beta = 8\alpha$ yields $R_{\beta} = R_0 (8\alpha)^{1/3} = 2 R_0 \alpha^{1/3}$. Dividing the two radii results in $\frac{R_{\alpha}}{R_{\beta}} = \frac{R_0 \alpha^{1/3}}{2 R_0 \alpha^{1/3}} = \frac{1}{2} = 0.5$.

Nuclear fusion releases energy because the product nucleus is more stable and possesses a higher total binding energy than the starting reactants. First, determine the total binding energy of the two deuterium nuclei, which contain four nucleons in total, by multiplying the nucleon count by the given binding energy per nucleon: $4 \times 1.1\ \mathrm{MeV} = 4.4\ \mathrm{MeV}$. Next, calculate the total binding energy of the resulting helium-4 nucleus, which has four nucleons at 7.0 MeV per nucleon, giving $4 \times 7.0\ \mathrm{MeV} = 28.0\ \mathrm{MeV}$. The net energy released during the fusion process is the difference between the final and initial total binding energies:
$Q = 28.0\ \mathrm{MeV} - 4.4\ \mathrm{MeV} = 23.6\ \mathrm{MeV}$

Bohr's atomic model establishes that the orbital radius of a hydrogen-like species is defined by the formula $r = \frac{a_0 n^2}{Z}$, where $n$ is the principal quantum number and $Z$ is the atomic number. For a $\mathrm{Li}^{++}$ ion, the atomic number is $Z=3$, and the ground state corresponds to $n=1$. Substituting these values into the model gives $r = \frac{a_0 (1)^2}{3}$, which simplifies to $r = \frac{1}{3} a_0$. Equating this to the form $\frac{1}{X} a_0$ reveals that $X=3$.

Nuclear density is a universal constant for all atomic nuclei, meaning the density of a copper nucleus is effectively the same as that of a carbon nucleus. Because the volume of a nucleus increases in direct proportion to its mass number $A$, the ratio of mass to volume remains unchanged regardless of the specific nucleus.

The expression $R \propto A^{1/3}$ accurately describes how nuclear radius scales with mass number, which is the underlying reason why nuclear density does not vary. Since this scaling law guarantees that density is independent of $A$, the claim that one element possesses a different nuclear density than another is incorrect, despite the scientific accuracy of the relationship mentioned in the reasoning.

The frequency of revolution of an electron is defined as the ratio of its orbital velocity to the circumference of its orbit, given by the relation $f = \frac{v}{2 \pi r}$. In Bohr's model, the electron's velocity is inversely proportional to the principal quantum number, $v \propto \frac{1}{n}$, while the radius of the orbit is proportional to the square of that number, $r \propto n^2$. Substituting these proportionalities into the frequency expression yields $f \propto \frac{1/n}{n^2}$, which simplifies to show that the frequency varies with $n$ as $\frac{1}{n^3}$.

For hydrogen-like ions, the energy of an electron is governed by the expression $E_n = -13.6 \frac{Z^2}{n^2}\text{ eV}$, where $n$ is the principal quantum number and $Z$ is the atomic number. The ground state is defined as $n = 1$, and since the first excitation state occurs at $n = 2$, the second excitation state corresponds to $n = 3$. The energy gap between these two states is calculated as $\Delta E = E_3 - E_1 = 13.6 Z^2 (1 - \frac{1}{9})$, which simplifies to $13.6 Z^2 \times \frac{8}{9}$. Setting this equal to 108.8 eV results in the equation $13.6 Z^2 \times \frac{8}{9} = 108.8$. Through algebraic simplification, $Z^2 = \frac{108.8 \times 9}{108.8}$, leading to $Z^2 = 9$ and an atomic number of 3.

The radioactive decay law $N(t) = N_0 e^{-\lambda t}$ describes the population of nuclei over time, where the decay constant for nucleus $n_2$ is triple that of nucleus $n_1$, meaning $\lambda_2 = 3\lambda$ and $\lambda_1 = \lambda$. Starting with the same initial quantity $N_0$ for both, the remaining number of each species after one half-life of $n_1$, defined by $t = \frac{\ln 2}{\lambda}$, is $N_1 = N_0 e^{-\lambda t}$ and $N_2 = N_0 e^{-3\lambda t}$. By dividing these expressions, the ratio simplifies to $\frac{N_2}{N_1} = \frac{e^{-3\lambda t}}{e^{-\lambda t}} = e^{-2\lambda (\frac{\ln 2}{\lambda})} = \frac{1}{4}$.

The observation that binding energy per nucleon remains nearly constant for nuclei with mass numbers between 30 and 170 is a fundamental consequence of the saturation effect inherent to nuclear forces. This behavior occurs precisely because the nuclear force is short-range, meaning each nucleon interacts only with its immediate neighbors rather than the entire nucleus, which prevents the binding energy from scaling indefinitely with size. Because the reason incorrectly labels nuclear forces as long-range, it stands in direct opposition to the established physical reality, confirming that the reason is false.

Bohr's atomic model determines the energy of single-electron species using the equation $E = -13.6 \frac{Z^{2}}{n^{2}} \text{ eV}$, where $Z$ is the atomic number and $n$ is the principal quantum number. A hydrogen atom \in its ground state ($Z=1, n=1$) possesses an energy of $-13.6 \text{ eV}$. For a $\mathrm{He}^{+}$ ion, $Z=2$, and its first excited state corresponds to $n=2$; substituting these values yields $E = -13.6 \frac{2^{2}}{2^{2}} = -13.6 \text{ eV}$, which confirms that statement (A) is correct. Similarly, a $\mathrm{Li}^{2+}$ ion has $Z=3$, and its second excited state occurs at $n=3$, resulting \in $E = -13.6 \frac{3^{2}}{3^{2}} = -13.6 \text{ eV}$, which proves statement (B) is also accurate. The remaining options involve energy states that do not result \in this same value of $-13.6 \text{ eV}$.

The energy released during an electron transition in a hydrogen atom is calculated by the difference between two energy levels using the formula $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ eV. An electron in the fourth excited state corresponds to an initial principal quantum number $n_i = 5$, and the problem requires finding the final state $n_f = n$ when the emitted photon energy is 2.86 eV. Substituting these known values into the equation yields $2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right)$, which rearranges to $\frac{2.86}{13.6} + \frac{1}{25} = \frac{1}{n^2}$. Calculating the decimal values gives approximately $0.21 + 0.04 = \frac{1}{n^2}$, simplifying to $0.25 = \frac{1}{n^2}$. Solving this equation results in $n^2 = 4$, which leads to an integer value of 2 for the final energy state.

Nuclear density represents the ratio of a nucleus's total mass to its physical volume. Since the mass of a nucleus is proportional to its mass number $A$ and its volume scales with $R^3$, substituting the relation $R = R_0 A^{1/3}$ into the volume formula results in the $A$ terms canceling out entirely. This calculation confirms that nuclear density is a constant property that does not change with the mass number.

The radius of an orbit for a hydrogen-like atom is given by the proportionality $r_n \propto \frac{n^2}{Z}$, where $n$ is the principal quantum number and $Z$ is the atomic number. For the $5^{\text{th}}$ orbit ($n=5$) of $\mathrm{Li}^{2+}$ ($Z=3$) and $\mathrm{He}^{+}$ ($Z=2$), the ratio of their radii is calculated as $\frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{n^2 / Z_{\mathrm{Li}}}{n^2 / Z_{\mathrm{He}}} = \frac{Z_{\mathrm{He}}}{Z_{\mathrm{Li}}} = \frac{2}{3}$.

Nuclear processes must adhere to strict conservation laws, particularly the principle that electric charge remains constant throughout the interaction. When a neutron decays into a proton, it must emit an electron to balance the net charge of the system. Since the creation of an electron is always physically paired with the emission of an antineutrino, the resulting nuclear process is $n \longrightarrow p + e^{-} + \overline{v}$.

Planck's constant $h$ is defined by the energy-frequency relationship $E = h\nu$, resulting in dimensions of $[ML^2T^{-1}]$, which are identical to the dimensions of angular momentum calculated as $L = mvr$ ([M] \cdot $[LT^{-1}]$ \cdot [L] = [ML^2T^{-1}]$). Bohr's quantization condition requires the angular momentum of an electron to be an integral multiple of the reduced Planck's constant, represented as$L = \frac{nh}{2\pi}$, meaning the statement claiming it is an integral multiple of$h$alone is physically inaccurate. Thus, the first statement is correct based on dimensional analysis, while the second statement is incorrect because it overlooks the necessary$2\pi$ factor in the quantization rule.

The Bohr model is restricted to hydrogen and hydrogen-like ions containing only a single electron because its mathematical framework assumes a central force arising exclusively from the electrostatic attraction between a nucleus of charge $+Ze$ and one orbiting electron. This derivation completely neglects the repulsive forces that would exist between multiple electrons, a factor which renders the theory inapplicable to any atom possessing more than one electron. Thus, because the model lacks the necessary parameters to account for inter-electronic repulsion, its scope is inherently limited to single-electron species, confirming the reason provided as the direct cause for the assertion.

In physics, any valid equation must satisfy dimensional consistency, meaning the physical dimensions on both sides of the equal sign must be identical. When analyzing the dimensions of the components, the term $p^2 c^2$ matches the dimensions of $E^2$, and the term $m^2 c^4$ also matches the dimensions of $E^2$. By verifying this dimensional homogeneity, it becomes clear that the correct relationship between energy, momentum, and mass is described by the equation $E^2=p^2 c^{2+}m^2 c^4$.

When an electron resides in an excited state, it can return to lower energy levels through various possible transitions, with each unique path producing a distinct spectral line. To determine the total number of these lines emitted when an electron drops from a higher level $n_2$ to a lower level $n_1$, we apply the formula:
$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
By substituting the given values of $n_2 = 4$ and $n_1 = 1$ into this expression, the calculation becomes:
$N = \frac{(4 - 1)(4 - 1 + 1)}{2} = \frac{3 \times 4}{2} = 6$
This confirms that the electron can undergo six different transition paths, resulting in a total of 6 spectral lines.