Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Radioactive decay follows an exponential progression described by the relation $A_t = A_0 e^{-\lambda t}$, where the decay constant $\lambda$ dictates the rate of reduction. The half-life, $T_{1/2}$, is defined as the time required for the activity to drop to half its initial value, which links to the decay constant via the fundamental expression $T_{1/2} = \frac{\ln 2}{\lambda}$.

To determine the half-life for this sample, we rearrange the decay law to $\lambda = \frac{1}{t} \ln \left( \frac{A_0}{A_t} \right)$. Given the activity changes from 700 to 500 in 30 minutes, we find the decay constant as $\lambda = \frac{\ln(700/500)}{30}$, or $\lambda = \frac{\ln(1.4)}{30}$. By substituting this value of $\lambda$ into the half-life formula, we get $T_{1/2} = \frac{30 \ln 2}{\ln(1.4)}$. Using the standard approximations $\ln 2 \approx 0.693$ and $\ln(1.4) \approx 0.336$, the calculation yields $T_{1/2} \approx \frac{0.693 \times 30}{0.336} \approx 61.8$ minutes, which rounds to 62 minutes.

The power output of a nuclear reactor is defined as the total energy released by the fission of fuel divided by the time taken for that fuel to be consumed. The total number of atoms in $2 \text{ kg}$ of $_{92}^{235}\mathrm{U}$ is calculated as $\frac{2 \times 6.023 \times 10^{26}}{235}$, and the energy released from each fission event is $200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}$. Multiplying these values yields the total energy produced, which must then be divided by the total time in seconds, given by $30 \times 24 \times 3600 \text{ s}$. This calculation results in a power output of approximately $63.2 \text{ MW}$, which is closest to the value of $60 \text{ MW}$.

The ionization energy of a hydrogen-like ion is governed by the expression $E = Z^2 \times 13.6 \text{ eV}$, where $Z$ is the atomic number. Since the provided ionization energy is 9 Rydbergs, and given that 1 Rydberg equals 13.6 eV, we equate $Z^2 \times 13.6 \text{ eV}$ to $9 \times 13.6 \text{ eV}$, which simplifies to $Z^2 = 9$, identifying the atomic number as $Z = 3$.

To find the wavelength of the radiation emitted during an electronic transition, we employ the Rydberg formula:
$\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
The electron transitions from the second excited state, where $n_2 = 3$, to the ground state, where $n_1 = 1$. Substituting the values $R = 1.09 \times 10^7 \text{ m}^{-1}$, $Z = 3$, $n_1 = 1$, and $n_2 = 3$ into the equation results in:
$\frac{1}{\lambda} = (1.09 \times 10^7 \text{ m}^{-1}) \times 3^2 \times \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = (1.09 \times 10^7 \text{ m}^{-1}) \times 9 \times \frac{8}{9}$
The factors of 9 cancel out, simplifying the expression to $\frac{1}{\lambda} = 8 \times 1.09 \times 10^7 \text{ m}^{-1} \approx 8.72 \times 10^7 \text{ m}^{-1}$. Calculating the reciprocal of this value yields a wavelength of approximately $1.14 \times 10^{-8} \text{ m}$, which is 11.4 nm.

The spectral lines of a hydrogen atom are defined by the Rydberg formula, where the wavelength depends on the transition between energy levels $n_2$ and $n_1$ according to the expression
$\lambda = \frac{c}{\frac{1}{n_1^2} - \frac{1}{n_2^2}}$
where $c$ is the inverse of the Rydberg constant. For the Lyman series, which involves transitions to the ground state $n_1=1$, the shortest wavelength occurs at the series limit $n_2=\infty$, yielding $\lambda_1 = c$, while the longest wavelength comes from the first transition $n_2=2$, resulting in $\lambda_2 = \frac{4c}{3}$. The difference between these wavelengths is $\frac{c}{3}$, which is given as $304 \text{\AA}$, establishing that $c = 912 \text{\AA}$.

When applying this same logic to the Paschen series, transitions land on $n_1=3$. The shortest wavelength occurs at $n_2=\infty$, giving $\lambda_1 = 9c$, and the longest wavelength occurs at $n_2=4$, calculated as $\lambda_2 = \frac{144c}{7}$. The difference for the Paschen series is therefore
$\Delta \lambda = \frac{144c}{7} - 9c = \frac{81c}{7}$
Substituting $c = 912 \text{\AA}$ into this result leads to $\frac{81 \times 912}{7}$, which equals $10553.14 \text{\AA}$.

The collision is governed by the conservation of linear momentum and total energy. Since the incident particle comes to a complete halt, its initial momentum $p$ must be transferred entirely to the recoiling hydrogen atom. We represent the initial kinetic energy of the particle as $K = \frac{p^2}{2m}$ and the recoil kinetic energy of the atom as $\frac{p^2}{2M}$, where $m = 200 \text{ MeV}/c^2$ and $M = 1 \text{ GeV}/c^2 = 1000 \text{ MeV}/c^2$. The conservation of energy requires that the particle's initial kinetic energy equals the sum of the atom's recoil energy and the $10.2 \text{ eV}$ excitation energy, written as $K = \frac{p^2}{2M} + 10.2$. By substituting the relationship $\frac{p^2}{2M} = K \frac{m}{M}$ into the energy balance equation, we obtain $K = K \frac{m}{M} + 10.2$, which simplifies to $K \left(1 - \frac{m}{M}\right) = 10.2$. Substituting the mass values, where $\frac{m}{M} = \frac{200}{1000} = 0.2$, leads to $0.8K = 10.2$, yielding $K = 12.75 \text{ eV}$. Finally, solving $\frac{N}{4} = 12.75$ gives $N = 51.00$.

For a nuclear process to be physically allowed, the total rest mass of the reactants must be greater than the total rest mass of the products, providing the necessary mass defect to satisfy the conservation of energy and momentum. In the reaction $n + p \rightarrow d + \gamma$, the reactant mass sum of 1.0087 u and 1.0072 u totals 2.0159 u, which is larger than the 2.0141 u mass of the deuteron product while the photon adds no mass. This condition of higher reactant mass ensures that the process is energetically favorable and adheres to the required conservation laws.

The density of a nucleus is defined as the ratio of its mass to its volume, a property that is surprisingly constant for all nuclei because both mass and volume scale linearly with the mass number $A$. The total mass of a nucleus is given by $M = A \cdot m_{\text{nucleon}}$, where $m_{\text{nucleon}} \cong 1.67 \times 10^{-27}$ kg, and its volume is determined by the sphere formula $V = \frac{4}{3} \pi R^3$. By substituting the given radius formula $R = (1.3 \times 10^{-15}) A^{1/3}$ m into the volume expression, we get $V = \frac{4}{3} \pi ((1.3 \times 10^{-15}) A^{1/3})^3$, which simplifies to $V = \frac{4}{3} \pi (1.3 \times 10^{-15})^3 A$. When calculating density, the mass number $A$ appears in both the numerator and the denominator and therefore cancels out, leading to the expression:

$\rho = \frac{m_{\text{nucleon}}}{\frac{4}{3} \pi (1.3 \times 10^{-15})^3}$

Substituting the provided value for the nucleon mass into this equation results in a nuclear density of approximately $2.3 \times 10^{17}$ kg m$^{-3}$. This calculation confirms that the order of magnitude for the density of a nucleus is $10^{17}$ kg m$^{-3}$.

The energy liberated in the nuclear reaction ${}_{3}^{7}\mathrm{Li} + {}_{1}^{1}\mathrm{H} \rightarrow 2({}_{2}^{4}\mathrm{He})$ is determined by Einstein's mass-energy equivalence principle $\Delta E = (\Delta m)c^2$, where the mass defect $\Delta m$ is the difference between the total mass of the reactants and the products. The mass defect is calculated as $\Delta m = (7.0160\,\mathrm{u} + 1.0079\,\mathrm{u}) - (2 \times 4.0026\,\mathrm{u}) = 8.0239\,\mathrm{u} - 8.0052\,\mathrm{u} = 0.0187\,\mathrm{u}$.

Using the standard conversion factor $1\,\mathrm{u} \approx 931.5\,\mathrm{MeV}/c^2$, the energy released per reaction is $\Delta E \approx 0.0187\,\mathrm{u} \times 931.5\,\mathrm{MeV}/\mathrm{u} \approx 17.419\,\mathrm{MeV}$. The number of lithium nuclei $N$ in $20\,\mathrm{g}$ of ${}_{3}^{7}\mathrm{Li}$ is found using the molar mass of $7\,\mathrm{g/mol}$ and Avogadro's number $N_A \approx 6.023 \times 10^{23}\,\mathrm{mol}^{-1}$, resulting in $N = (\frac{20}{7}) \times 6.023 \times 10^{23} \approx 1.72 \times 10^{24}$.

The total energy released in Joules is $E_{\text{total}} = N \times \Delta E \times (1.6 \times 10^{-13}\,\mathrm{J}/\mathrm{MeV}) \approx (1.72 \times 10^{24}) \times 17.419 \times 1.6 \times 10^{-13}\,\mathrm{J} \approx 4.79 \times 10^{12}\,\mathrm{J}$. Converting this value to kilowatt-hours using the factor $1\,\mathrm{kWh} = 3.6 \times 10^6\,\mathrm{J}$ gives $E = \frac{4.79 \times 10^{12}}{3.6 \times 10^6} \approx 1.33 \times 10^6\,\mathrm{kWh}$.

In the Rutherford gold foil experiment, the distribution of scattered alpha particles follows the relationship $Y \propto \frac{1}{\sin^4(\theta/2)}$ where $Y$ denotes the detected particle count and $\theta$ is the scattering angle. Since the number of particles is inversely proportional to the fourth power of the sine of half the scattering angle, the intensity falls off extremely rapidly as the angle increases from zero. This mathematical behavior results in a steep, continuous curve that characteristically drops sharply and approaches the axis as the scattering angle grows larger.

Radioactive decay is governed by the exponential relation $R = R_{0} e^{-\lambda t}$, where the decay constant $\lambda$ represents the rate at which a substance loses activity over time. Since the half-life of a substance is defined by the formula $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$, it is inversely proportional to the decay constant derived from the slope of each activity curve. By inspecting the slopes of the provided graphs, we identify the decay constants as $\lambda_{A} = \frac{6}{10}$, $\lambda_{B} = \frac{6}{5}$, and $\lambda_{C} = \frac{2}{5}$. Calculating the respective half-lives gives $T_{A} = \frac{10}{6} \ln 2$, $T_{B} = \frac{5}{6} \ln 2$, and $T_{C} = \frac{5}{2} \ln 2$. Expressing these values with a common denominator leads to $T_{A} = \frac{10}{6} \ln 2$, $T_{B} = \frac{5}{6} \ln 2$, and $T_{C} = \frac{15}{6} \ln 2$, which simplifies to a ratio of $2:1:3$.

The stability of a nucleus is determined by the energy released during its formation from individual protons and neutrons, known as binding energy. This process involves a mass deficit, where the total mass of the constituent particles exceeds the actual mass of the resulting nucleus, with the difference converting into energy according to the principle of mass-energy equivalence.

Calculating this value for tin requires determining the expected mass by combining the masses of 50 protons and 70 neutrons.
$M_{\text{expected}} = 50(1.00783) + 70(1.00867) = 120.9984 \text{ U}$
Subtracting the actual mass of 119.902199 U from this expected mass reveals a mass defect of 1.096201 U. Converting this mass difference into energy using the provided factor of 931 MeV per U results in a total binding energy of 1020.56 MeV. Dividing this energy by the total number of nucleons, 120, produces a binding energy per nucleon of 8.5 MeV.

The Rydberg formula establishes the wavelength of emitted radiation in hydrogen atoms using the relationship
$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
where $n_1$ and $n_2$ represent the lower and higher energy levels of the transition. For the Balmer series, the electron always transitions to $n_1 = 2$, while the first and second members occur at $n_2 = 3$ and $n_2 = 4$, respectively.

Because the Rydberg constant $R_H$ is common to both transitions, comparing two wavelengths within the same series cancels this constant out, allowing for a direct ratio calculation:
$\frac{\lambda_2}{\lambda_1} = \frac{(\frac{1}{2^2} - \frac{1}{3^2})}{(\frac{1}{2^2} - \frac{1}{4^2})}$
Evaluating these fractions simplifies the expression to the ratio of $5/36$ to $3/16$, meaning the second wavelength is found by calculating $\lambda_2 = 6561 \text{ \AA} \times \frac{20}{27}$. This operation yields 4860 \AA, which is equivalent to 486 nm.

When a radioactive nucleus decays via multiple competing pathways, the total decay rate is the sum of the rates for each individual process. Because the decay constant is inversely proportional to the half-life, the effective half-life $T_{\text{eff}}$ is defined by the following relationship:

$\frac{1}{T_{\text{eff}}} = \frac{1}{T_1} + \frac{1}{T_2}$

Substituting the given values of 10 s and 100 s into this expression allows us to calculate the effective half-life as:

$T_{\text{eff}} = \frac{T_1 T_2}{T_1 + T_2} = \frac{10 \cdot 100}{10 + 100} = \frac{1000}{110}$

This calculation results in an effective half-life of approximately 9 s.

The energy gained by a charged particle moving through an electric potential difference is described by the relation $q \Delta V = \frac{1}{2} m v^2$, which signifies that the electrical work performed on the particle directly converts to its kinetic energy. By rearranging this expression to isolate velocity, we obtain $v = \sqrt{\frac{2 q \Delta V}{m}}$, revealing that the speed of the particle is inversely proportional to the square root of its mass when the charge and potential difference remain constant. Since both the hydrogen ion and the singly ionized helium atom possess an identical charge magnitude of $e$, their speed ratio depends entirely on the inverse square root of their masses. Given that a helium ion is approximately four times as massive as a hydrogen ion, the ratio of the hydrogen speed to the helium speed is $\sqrt{\frac{4m}{m}}$, which simplifies to 2.

The energy levels of a hydrogen atom are determined by the principal quantum number $n$ according to the expression $E_n = -\frac{E_0}{n^2}$, where $E_0$ is the ionization energy. When an electron transitions from the $(n+1)^{\text{th}}$ level to the $n^{\text{th}}$ level, the emitted photon carries an energy equal to the difference between these levels, given by $\Delta E = E_0 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$, which simplifies to $\Delta E = E_0 \left( \frac{2n+1}{n^2(n+1)^2} \right)$. In the regime where the quantum number is very large, specifically $n \gg 1$, the term $(n+1)$ is approximately equal to $n$, allowing the energy difference to be approximated as $E_0 \left( \frac{2n}{n^2 \cdot n^2} \right)$, which simplifies further to $\frac{2E_0}{n^3}$. Since the energy of the emitted radiation is related to its frequency $\nu$ by $E = h\nu$, it follows that the frequency is proportional to $\frac{1}{n^3}$.

Radioactive decay follows the first-order exponential relation $N(t) = N_0 e^{-\lambda t}$, where the fraction of active material remaining is expressed as $e^{-\lambda t}$. The problem states that this fraction equals $\frac{9}{16}$ at time $t$. To determine the fraction remaining after $\frac{t}{2}$, we evaluate the expression $e^{-\lambda (t/2)}$, which is mathematically equivalent to the square root of the original fraction $e^{-\lambda t}$. Taking the square root of $\frac{9}{16}$ results in $\frac{3}{4}$ as the fraction of the material remaining.

The time period of an electron in a hydrogen atom is directly proportional to the cube of the principal quantum number $n$, given by the relationship $T \propto n^3$. Since the first excited state corresponds to $n=2$ and the ground state corresponds to $n=1$, the time period increases by a factor of 8. Using the provided ground state time period of $1.6 \times 10^{-16}$ s, the new time period is $8 \times 1.6 \times 10^{-16}$ s. The frequency is the reciprocal of the time period, leading to the calculation:

$f = \frac{1}{8 \times 1.6 \times 10^{-16}} \approx 7.8 \times 10^{14} \text{ s}^{-1}$

The nuclear mass density is a universal constant because the volume of a nucleus increases in direct proportion to its mass number. Given that the nuclear radius scales as $R = R_{0} A^{1/3}$, the volume, which depends on the cube of the radius, becomes proportional to $A$, thereby cancelling the mass number when calculating density. As a result, the density remains uniform for all nuclei, causing the ratio between $^{40}\mathrm{Ca}$ and $^{16}\mathrm{O}$ to be 1.

Radioactive activity follows the fundamental relationship $A = \lambda N$, where activity is directly proportional to the decay constant $\lambda$ and the number of nuclei $N$. For sample A, we write $10 = \lambda_A(2N_0)$, and for sample B, we write $20 = \lambda_B(N_0)$. Dividing these expressions confirms $\lambda_B = 4\lambda_A$, and because the half-life $T_{1/2} = \frac{\ln 2}{\lambda}$ is inversely proportional to the decay constant, we establish $(T_{1/2})_A = 4(T_{1/2})_B$. This inverse relationship is satisfied when sample A has a half-life of 20 days and sample B has a half-life of 5 days.

The behavior of an electron transitioning between energy levels in a hydrogen atom is governed by the Rydberg formula, which relates the wavelength of the emitted photon to the principal quantum numbers of the final and initial states. When an electron descends from the third excited state, $n=4$, to the first excited state, $n=2$, it emits photons during a two-step cascade, first dropping from $n=4$ to $n=3$ and then from $n=3$ to $n=2$. Applying the formula $\frac{1}{\lambda}=R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, the inverse wavelength for the first transition is $\frac{1}{\lambda_1}=R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = \frac{7R}{144}$, while for the second transition it is $\frac{1}{\lambda_2}=R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5R}{36}$. Taking the ratio of these wavelengths yields $\frac{\lambda_1}{\lambda_2} = \frac{144}{7R} \times \frac{5R}{36} = \frac{20}{7}$.