Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

The wavelength of radiation emitted by a hydrogen atom is governed by the Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For the Lyman series, where $n_1=1$, the maximum wavelength corresponds to the transition from $n_2=2$, yielding $\frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$, which simplifies to $\lambda = \frac{4}{3R}$. The Balmer series, defined by transitions ending at $n_1=2$, is known to lie within the visible region of the electromagnetic spectrum. For the Paschen series, where $n_1=3$, the minimum wavelength occurs at the limit $n_2 \to \infty$, resulting in $\frac{1}{\lambda} = R \left( \frac{1}{9} - 0 \right) = \frac{R}{9}$, so $\lambda = \frac{9}{R}$. Conversely, the minimum wavelength of the Lyman series is calculated with $n_2 \to \infty$ to be $\lambda = \frac{1}{R}$, confirming that the value $\frac{5}{4R}$ in the final statement is incorrect and validating statements A, B, and C.

The orbital radius in a hydrogen atom is proportional to the square of the principal quantum number $n$, allowing the transition states to be determined by the ratio $\frac{r_i}{r_f} = \frac{n_i^2}{n_f^2}$. Given the provided ratio of 16 to 4, the square of the ratio of the quantum numbers is 4, meaning $\frac{n_i}{n_f} = 2$, which identifies the transition as occurring from the fourth energy level down to the second energy level.

The wavelength of the emitted photon is determined using the Rydberg equation $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. Substituting the Rydberg constant $1.0973 \times 10^7 \text{ /m}$ and the quantum numbers $n_f = 2$ and $n_i = 4$ results in $\frac{1}{\lambda} = 1.0973 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 1.0973 \times 10^7 \left( \frac{3}{16} \right)$. Solving for $\lambda$ yields approximately $4.86 \times 10^{-7} \text{ m}$, which is equivalent to 486 nm.

In Rutherford's experiment, the atom is characterized by its vast empty space, with virtually all its mass and positive charge concentrated within a nucleus that is miniscule in size compared to the entire atom. Since the cross-sectional area of this nucleus is so small, an alpha particle is statistically unlikely to encounter it, which explains why only a very small fraction of particles experience a head-on collision. These few particles that do rebound are precisely the ones that approach the nucleus with an impact parameter $b$ approaching zero, resulting in a direct interaction that causes the alpha particle to reverse its path due to strong Coulombic repulsion.

In spontaneous $\alpha$-decay, the parent nucleus at rest must disintegrate into an $\alpha$-particle and a daughter nucleus, with the total energy $Q$ being shared between the two according to the conservation of linear momentum. This physical law requires the kinetic energy $K_{\alpha}$ of the emitted $\alpha$-particle to be a specific fraction of the total $Q$ value, determined by the mass number $A$ of the parent nucleus as $K_{\alpha} = \left( \frac{A-4}{A} \right) Q$. Given that both radioactive substances possess an identical $Q$ value of 1 MeV, their respective $\alpha$-particle energies are simply scaled by their parent mass numbers. Calculating these energies for parent mass numbers 200 and 212 results in the ratio of the two energies determined by the following expression:

$\frac{K_{\alpha_A}}{K_{\alpha_B}} = \left( \frac{196}{200} \right) \times \left( \frac{212}{208} \right) = \frac{2597}{2600}$

The Rydberg formula for the wavelength of hydrogen spectral lines is $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. For the Lyman series limit, $n_f = 1$ and $n_i = \infty$, so $\frac{1}{\lambda_{L,min}} = R$, resulting in $R = \frac{1}{91} \text{ nm}^{-1}$. The largest wavelength of the Paschen series corresponds to the transition from $n_i = 4$ to $n_f = 3$, giving $\lambda_{P,max} = \frac{1}{R (\frac{1}{3^2} - \frac{1}{4^2})} = \frac{144}{7} \times 91 = 1872 \text{ nm}$. The largest wavelength of the Balmer series occurs for the transition from $n_i = 3$ to $n_f = 2$, yielding $\lambda_{B,max} = \frac{1}{R (\frac{1}{2^2} - \frac{1}{3^2})} = \frac{36}{5} \times 91 = 655.2 \text{ nm}$. Subtracting these values provides a difference of $1872 \text{ nm} - 655.2 \text{ nm} = 1216.8 \text{ nm}$, which is approximately $1217 \text{ nm}$.

The radius $r_n$ and velocity $v_n$ of an electron in a hydrogen-like atom are related to the orbit number $n$ and atomic number $Z$ by $r_n \propto \frac{n^2}{Z}$ and $v_n \propto \frac{Z}{n}$. Equating the radii $r_1 = r_2$ gives the relationship $\frac{n_1^2}{Z_1} = \frac{n_2^2}{Z_2}$, which implies $\frac{Z_1}{Z_2} = \left(\frac{n_1}{n_2}\right)^2$. The ratio of the speeds is given by $\frac{v_1}{v_2} = \frac{Z_1/n_1}{Z_2/n_2} = \frac{Z_1}{Z_2} \times \frac{n_2}{n_1} = 1.2$. Substituting the atomic number ratio into this expression yields $\left(\frac{n_1}{n_2}\right)^2 \times \frac{n_2}{n_1} = \frac{n_1}{n_2} = 1.2 = \frac{6}{5}$. Thus, the possible energy states are $n_1 = 6$ and $n_2 = 5$.

The energy of an electron in the $n$-th orbit is related to the ground state energy $E_0$ by the formula $E_n = \frac{E_0}{n^2}$. Setting the magnitude $|E_n| = 0.04 |E_0|$ results in the equation $\frac{1}{n^2} = 0.04$, which simplifies to $n^2 = 25$ and yields $n = 5$. According to Bohr's quantization condition, the angular momentum is $L = \frac{nh}{2\pi}$, which rearranges to $\frac{2 \pi L}{h} = n$. Substituting $n = 5$ into this expression gives the result 5.

The momentum of a photon is inversely related to its wavelength through the de Broglie relation $p = \frac{h}{\lambda}$, which implies that finding the ratio of momenta for two spectral lines is equivalent to determining the inverse ratio of their wavelengths. The Balmer series originates from electron transitions falling to the $n=2$ energy level of a hydrogen atom. Applying the Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, the wavelength of the first line, corresponding to the transition from $n=3$ to $n=2$, is found using $\frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} \right)$, resulting in $\frac{1}{\lambda_1} = \frac{5R}{36}$, or $\lambda_1 = \frac{36}{5R}$. For the second line, which corresponds to the transition from $n=4$ to $n=2$, the calculation $\frac{1}{\lambda_2} = R \left( \frac{1}{4} - \frac{1}{16} \right)$ yields $\frac{1}{\lambda_2} = \frac{3R}{16}$, or $\lambda_2 = \frac{16}{3R}$. By taking the ratio of the momenta $\frac{p_1}{p_2} = \frac{\lambda_2}{\lambda_1}$, we substitute the derived wavelengths to obtain $\frac{\frac{16}{3R}}{\frac{36}{5R}}$, which simplifies to $\frac{16}{3} \times \frac{5}{36} = \frac{80}{108}$. This fraction reduces to $\frac{20}{27}$, identifying $\alpha$ as 20 and $\beta$ as 27.

The Einstein mass-energy equivalence principle states that the energy of the photon required to initiate the reaction must equal the energy equivalent of the mass defect, defined by the relation $E = \Delta m c^2 = h\nu$. The mass defect $\Delta m$ is the difference between the total mass of the products and the mass of the reactant, calculated as $\Delta m = (4 \times 4.002) - 15.348 = 16.008 - 15.348 = 0.66 \text{ amu}$. Converting this mass defect to kilograms using the given conversion factor yields $\Delta m = 0.66 \times 1.66 \times 10^{-27} \text{ kg} = 1.0956 \times 10^{-27} \text{ kg}$.

Substituting these values into the frequency equation $\nu = \frac{\Delta m c^2}{h}$, the calculation becomes:
$\nu = \frac{1.0956 \times 10^{-27} \times (3 \times 10^8)^2}{6.6 \times 10^{-34}} = \frac{1.0956 \times 10^{-27} \times 9 \times 10^{16}}{6.6 \times 10^{-34}} = 1.494 \times 10^{23} \text{ Hz}$
Converting this result into kilohertz by dividing by $10^3$, the frequency is $1.494 \times 10^{20} \text{ kHz}$, which is equivalent to $14.94 \times 10^{19} \text{ kHz}$.

When an electron revolves in a circular orbit within a hydrogen atom, it effectively creates a current loop, generating a magnetic field at the center given by:
$B = \frac{\mu_0 I}{2r}$
The circulating electron with charge $e$ constitutes an effective current $I = \frac{ev}{2\pi r}$, where $v$ is the orbital velocity and $r$ is the orbital radius. Substituting this current into the magnetic field expression leads to the relation $B = \frac{\mu_0 e v}{4\pi r^2}$. From the Bohr model, the orbital radius follows the relation $r \propto n^2$ while the orbital velocity follows $v \propto \frac{1}{n}$. Consequently, the magnetic field depends on the principal quantum number $n$ as $B \propto \frac{(1/n)}{(n^2)^2}$, which simplifies to the proportionality $B \propto \frac{1}{n^5}$. Therefore, calculating the ratio of the magnetic fields for the second and fourth orbits results in:
$\frac{B_2}{B_4} = \left( \frac{4}{2} \right)^5 = 2^5 = 32$

Nuclear energy release is determined by the difference in total binding energy between the reactants and the product, where total binding energy is calculated by multiplying the mass number by the binding energy per nucleon. To find the initial binding energy, sum the contributions of the two mass-3 nuclei and the single mass-4 nucleus: $2 \times 3 \times 5.6 + 4 \times 7.4 = 33.6 + 29.6 = 63.2 \text{ MeV}$. The single mass-10 product nucleus possesses a total binding energy of $10 \times 6.1 = 61 \text{ MeV}$. The energy change $\Delta M c^2$ is the absolute difference between these initial and final states, calculated as $|61 - 63.2| = 2.2 \text{ MeV}$.

The mass defect of a nucleus is the difference between the combined mass of its constituent nucleons and the atom's actual experimental mass. For a carbon-12 atom, this calculation requires accounting for the sum of the masses of 6 protons and 6 neutrons compared to its defined atomic mass of 12 u.

To determine this value, we use the formula $\Delta m = 6(1.00727 + 1.00866) - 12$. Adding the masses of the individual nucleons within the parentheses gives $6(2.01593)$, which equals 12.09558. Subtracting the atomic mass of 12 u leaves a mass defect of 0.09558 u. To convert this result into energy, we multiply by the given factor of 931.5 MeV/c², where $0.09558 \times 931.5$ yields 89.03 MeV/c².

The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$, which implies the surface area of a nucleus $S = 4\pi R^2$ is proportional to $A^{2/3}$. For the initial atom ${}_3^8 X$, the mass number $A_i$ is 8, and after absorbing 10 protons and 9 neutrons, the new mass number becomes $A_f = 8 + 10 + 9 = 27$, as electrons do not contribute to the nuclear structure.

The initial surface area is proportional to $8^{2/3} = 4$, while the final surface area is proportional to $27^{2/3} = 9$. The percentage growth is calculated as:
$\frac{S_f - S_i}{S_i} \times 100\% = \frac{9 - 4}{4} \times 100\% = 125\%$

The fundamental principle behind this nuclear transformation is Einstein's mass-energy equivalence, which dictates that the mass lost during a reaction manifests as released energy. For the reaction involving lithium-7 and a proton, written as $^{7}_{3}\text{Li} + ^{1}_{1}\text{p} \rightarrow 2^{4}_{2}\text{He}$, the mass defect $\Delta m$ is calculated by subtracting the total final mass from the total initial mass:
$\Delta m = (7.0183 + 1.008) - 2(4.004) = 0.0183 \text{ u}$
Since 1 u corresponds to 931 MeV, the energy released in each individual reaction is $0.0183 \times 931 \times 10^6$ eV.

The total energy released depends on the number of lithium atoms in the sample. A mass of 7/17.13 kg is equivalent to 7000/17.13 g. Given the molar mass of lithium is 7 g/mol, the number of moles is $1000 / 17.13$, and multiplying by the Avogadro number $6.0 \times 10^{23}$ gives the total number of nuclei as $6 \times 10^{26} / 17.13$. The total energy is the product of the number of reactions and the energy released per reaction:
$E_{\text{total}} = \left( \frac{6 \times 10^{26}}{17.13} \right) \times (0.0183 \times 931 \times 10^6) \text{ eV}$
Evaluating this expression yields $E_{\text{total}} \approx 6 \times 10^{32}$ eV, which indicates the value of $\alpha$ is 6.

The binding energy per nucleon represents the average energy holding each proton and neutron together within an atomic nucleus, determined by calculating the mass defect and converting it into energy. For the nucleus ${}_{83}^{209}\mathrm{Bi}$, the mass defect arises because the combined mass of the 83 individual protons and 126 individual neutrons exceeds the actual mass of the bismuth nucleus. This mass difference is calculated using the equation $\Delta m = 83(1.007825\,\text{u}) + 126(1.008665\,\text{u}) - 208.980388\,\text{u} = 1.760877\,\text{u}$. Converting this mass defect into binding energy using the factor $931\,\text{MeV}$ per atomic mass unit results \in a total binding energy of $BE = 1.760877 \times 931\,\text{MeV} = 1639.3764\,\text{MeV}$. Finally, dividing this total by the 209 nucleons leads to an average binding energy per nucleon of $\frac{BE}{A} = \frac{1639.3764\,\text{MeV}}{209} \approx 7.84\,\text{MeV}$.

The distance of closest approach $r_0$ for an $\alpha$-particle scattering off a nucleus is determined by equating the initial kinetic energy $E$ of the $\alpha$-particle to the electrostatic potential energy at the moment the particle stops and reverses direction: $E = \frac{1}{4\pi \epsilon_0} \frac{(2e)(Ze)}{r_0}$. Substituting the values $E = 7.9 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}$, $Z = 79$, and the Coulomb constant $k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$, the expression for the distance of closest approach is $r_0 = \frac{k(2e)(Ze)}{E}$. Calculating this results in $r_0 = 2.88 \times 10^{-14} \text{ m}$. Since the diameter of the target nucleus is defined in this context as twice the distance of closest approach, the estimated diameter is $2 \times 2.88 \times 10^{-14} \text{ m} = 5.76 \times 10^{-14} \text{ m}$.

The binding energy per nucleon does not monotonically increase with the mass number of the nucleus, as the binding energy curve rises to a maximum for intermediate-mass elements like iron ($^{56}\text{Fe}$) and subsequently decreases for heavier nuclei, which renders Statement I false. In nuclear processes, atomic systems inherently move toward greater stability, characterized by higher binding energy per nucleon, which is the physical principle driving both the fusion of light nuclei and the fission of heavy nuclei toward a more stable configuration, thereby confirming that Statement II is true.

The stability of a nucleus in the context of these reactions is directly proportional to its total binding energy, as a higher binding energy indicates a more stable configuration relative to the reactants. Let $B_3, B_4,$ and $B_5$ represent the binding energies of ${ }_2 He ^3, { }_2 He ^4,$ and ${ }_2 He ^5$, respectively. From the first reaction, the energy balance is given by $B_4 - B_3 = 20 \text{ MeV}$, which implies $B_4 = B_3 + 20 \text{ MeV}$. From the second reaction, the balance is $B_5 - B_4 = -0.9 \text{ MeV}$, which can be expressed in terms of $B_3$ as $B_5 = B_4 - 0.9 \text{ MeV} = (B_3 + 20 \text{ MeV}) - 0.9 \text{ MeV} = B_3 + 19.1 \text{ MeV}$. Comparing the expressions for these energies shows that $B_4 > B_5 > B_3$, and consequently, the stability order is $X_4 > X_5 > X_3$.

Bohr's quantization condition defines the angular momentum of an electron as $L = \frac{nh}{2\pi}$. Setting this equal to the provided value of $\frac{3h}{\pi}$ allows us to determine the principal quantum number by solving the equation $\frac{nh}{2\pi} = \frac{3h}{\pi}$, which simplifies to $n = 6$. The energy of the electron is then calculated using the hydrogen atom energy formula $E = \frac{-13.6}{n^2}$, which yields $E = \frac{-13.6}{6^2}$ and results in a final value of -0.38 eV.

Isobars are defined as nuclides that possess the same mass number ($A$) but different atomic numbers ($Z$). The pair ${ }_1^3 H$ and ${ }_2^3 He$ qualifies as isobars because both nuclei have an identical mass number of 3, while their atomic numbers are 1 and 2, respectively. The other provided pairs are either isotopes, sharing the same atomic number, or exhibit different mass numbers entirely, failing to meet the definition of isobars.