Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

The volume $V$ of blood is determined using the relation $V = \frac{A_t}{A_{unit}}$, where $A_t$ is the total activity remaining after time $t$ and $A_{unit}$ is the activity per cubic centimeter. The initial activity $A_0$ is $0.8 \times 10^{-6} \times 3.7 \times 10^{10} = 29600$ decays per second. The activity remaining after 10 hours is calculated as $A_t = A_0 e^{-\lambda t} = 29600 \times 0.84 = 24864$ decays per second. The activity per unit volume $A_{unit}$ is $\frac{300}{60} = 5$ decays per second per cubic centimeter. Finally, the total volume is $V = \frac{24864}{5} = 4972.8$ cubic centimeters, which is approximately 5 liters.

When a neutron undergoes a head-on elastic collision with a stationary target nucleus, the fractional loss of its initial kinetic energy depends entirely on the masses of the two particles. The general expression for this fractional energy loss is given by $\frac{\Delta K}{K} = \frac{4m_1m_2}{(m_1+m_2)^2}$, where $m_1$ represents the mass of the neutron and $m_2$ represents the mass of the stationary nucleus.

Taking the mass of the neutron as 1 unit, we can determine the energy loss for each target by substituting its relative mass into the formula. For deuterium, which has a mass of 2, the fractional loss is $\frac{4(1)(2)}{(1+2)^2} = \frac{8}{9}$, which calculates to 0.89. Similarly, for the carbon nucleus with a mass of 12, the fractional loss becomes $\frac{4(1)(12)}{(1+12)^2} = \frac{48}{169}$, which simplifies to approximately 0.28.

The Rydberg formula establishes that the frequency of spectral lines depends on the energy levels involved, where the series limit corresponds to a transition from infinity to the specific starting principal quantum number $n_1$. For the Lyman series, electrons drop to $n_1 = 1$, resulting in a frequency $v_L = cR(\frac{1}{1^2} - \frac{1}{\infty ^2}) = cR$. Similarly, the Pfund series involves transitions to the $n_1 = 5$ energy level, giving a frequency of $f = cR(\frac{1}{5^2} - \frac{1}{\infty ^2}) = \frac{cR}{25}$. Since $cR$ is equal to $v_L$, substituting this into the Pfund frequency expression yields $\frac{v_L}{25}$.

The energy required to remove an electron from a hydrogen-like species is determined by the formula $E = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$. For a singly ionized helium atom (\text{He}^+$) where the atomic number$Z=2$and the electron occupies the ground state$n=1$, the second ionization energy is calculated as$E_2 = 13.6 × \frac{2^2}{1^2} = 54.4 \text{ eV}$. Since this energy is defined as 2.2 \times the energy required to remove the first electron from a neutral helium atom ($E_1$), the first ionization energy is$E_1 = \frac{54.4}{2.2} \approx 24.73 \text{ eV}$. The total energy required to fully ionize the helium atom is the \sum of these two values, yielding$E_{total} = E_1 + E_2 = 24.73 + 54.4 \approx 79.13 \text{ eV}$, which rounds to 79 eV.

The energy of a photon emitted or absorbed during a transition between energy levels is inversely proportional to its wavelength, defined by the relationship $\Delta E = \frac{hc}{\lambda}$. To determine the ratio of the wavelengths, we calculate the energy difference for each specific transition shown. The transition for wavelength $\lambda_1$ occurs between the levels $-E$ and $-2E$, resulting in an energy difference of $\Delta E_1 = -E - (-2E) = E$, which gives the expression $\lambda_1 = \frac{hc}{E}$.

The transition for wavelength $\lambda_2$ takes place between the levels $-E$ and $-\frac{4E}{3}$, yielding an energy difference of $\Delta E_2 = -E - (-\frac{4E}{3}) = \frac{E}{3}$. Substituting this into the wavelength formula results in $\lambda_2 = \frac{hc}{E/3} = \frac{3hc}{E}$. By calculating the ratio of these two wavelengths, we find $r = \frac{\lambda_1}{\lambda_2} = \frac{hc/E}{3hc/E}$, which simplifies to $\frac{1}{3}$.

The centripetal acceleration of an electron moving in a circular orbit is given by $a = \frac{v^2}{r}$, where $v$ is the orbital velocity and $r$ is the radius of the orbit. According to Bohr's quantization condition for angular momentum, $mvr = \frac{nh}{2\pi}$, which for the first orbit ($n=1$) simplifies to $mvr = \frac{h}{2\pi}$. Solving this equation for the orbital velocity yields $v = \frac{h}{2\pi mr}$.

Substituting this expression for $v$ into the acceleration formula gives $a = \frac{(\frac{h}{2\pi mr})^2}{r}$. Squaring the term in the numerator results in $a = \frac{h^2}{4\pi^2 m^2 r^2 \cdot r}$, which simplifies to $a = \frac{h^2}{4\pi^2 m^2 r^3}$.

Radioactive decay follows the fundamental law $N_A = N_0 e^{-\lambda t}$, where $N_0$ represents the initial quantity of substance $A$, and the decay constant relates to half-life via $\lambda = \frac{\ln(2)}{T}$. Because the total number of nuclei remains conserved throughout the process, the sum of the remaining nuclei $N_A$ and the decayed nuclei $N_B$ equals the initial amount, $N_A + N_B = N_0$. Given the ratio $\frac{N_B}{N_A} = 0.3$, we can express $N_B$ as $0.3 N_A$, which leads to the relation $N_A + 0.3 N_A = 1.3 N_A = N_0$, or $N_A = \frac{N_0}{1.3}$.

Substituting this expression for $N_A$ into the decay law yields:
$\frac{N_0}{1.3} = N_0 e^{-\lambda t}$

Dividing both sides by $N_0$ gives $\frac{1}{1.3} = e^{-\lambda t}$, which simplifies to $1.3 = e^{\lambda t}$. Taking the natural logarithm of both sides results in:
$\ln(1.3) = \lambda t$

Replacing $\lambda$ with $\frac{\ln(2)}{T}$ allows us to solve for time $t$:
$t = \frac{\ln(1.3)}{\frac{\ln(2)}{T}} = T \frac{\log(1.3)}{\log(2)}$

Nuclear fusion releases energy when the product nucleus is more stable than the reactants, which occurs because the total binding energy of the final product exceeds that of the initial nuclei. The two reacting deuterons contribute a total of four nucleons, each with a binding energy of 1.1 MeV, yielding an initial total binding energy of $2 \times 2 \times 1.1 = 4.4$ MeV. The resulting helium nucleus contains four nucleons with a binding energy of 7.0 MeV, resulting in a final total binding energy of $4 \times 7.0 = 28$ MeV. Subtracting the initial total from the final total reveals the energy released in the reaction, calculated as $28 - 4.4 = 23.6$ MeV.

The mass-energy equivalence principle, defined by the equation $E = mc^2$, relates the energy produced to the mass consumed and the square of the speed of light. Since power is the rate of energy consumption ($P = \frac{E}{t}$), the total energy produced \in one hour is represented as $E = P \times t$, which allows the mass to be calculated using the formula $m = \frac{P \times t}{c^2}$. Substituting the given values where $P = 10^9$ Watts, $t = 3600$ seconds, and $c = 3 × 10^8$ m/s leads to $m = \frac{10^9 \times 3600}{(3 \times 10^8)^2}$. Simplifying this calculation results \in $m = \frac{3.6 \times 10^{12}}{9 \times 10^{16}} = 0.4 \times 10^{-4} \text{ kg}$. Converting this mass into grams by multiplying by $10^3$ gives $0.4 \times 10^{-1} \text{ g} = 4 \times 10^{-2} \text{ g}$.

The magnetic field generated at the center of a current-carrying circular loop is expressed by $B = \frac{\mu_0 i}{2r}$, showing that the field strength is directly proportional to the current and inversely proportional to the radius of the orbit. Within Bohr's model of a hydrogen atom, the radius of the nth orbit follows the relationship $r_n \propto n^2$, while the equivalent current created by the circulating electron scales as $i_n \propto n^{-3}$. Substituting these dependencies into the magnetic field equation leads to the proportionality $B \propto \frac{n^{-3}}{n^2}$, confirming that the magnetic field is proportional to $n^{-5}$.

The energy of a photon emitted during a transition between energy levels in a hydrogen atom is determined by the difference in energy between the initial and final states, which equates to 10.2 eV for a transition from level 2 to level 1. To completely remove an electron from a hydrogen-like ion, the incident photon must possess energy at least equal to the ionization energy of the ion in its specific state, given by the expression $13.6 \cdot \frac{Z^2}{n^2}$ eV. Given the lithium atomic number of 3, the binding energy required is $13.6 \cdot \frac{9}{n^2}$ eV, leading to the inequality $10.2 > \frac{13.6 \cdot 9}{n^2}$.

Rearranging this expression to isolate the squared quantum number gives $n^2 > \frac{13.6 \cdot 9}{10.2}$, which simplifies to $n^2 > 12$. Since the quantum number must be an integer, we seek the smallest whole number whose square exceeds 12. Because 3 squared is 9 and 4 squared is 16, the lowest quantum number required for this process is 4.

The radioactive decay process is fundamentally governed by the number of half-lives elapsed, which dictates the proportion of nuclei that remain versus those that have decayed. The number of decayed nuclei is calculated as the initial amount minus the remaining amount, expressed as $N_d = N_0(1 - (1/2)^n)$, where $n$ represents the number of half-lives $n = t/T_{1/2}$.

For element A, the total duration of 80 minutes over a half-life of 20 minutes results in $n = 4$ cycles. Consequently, the number of decayed nuclei for A is $N_0(1 - (1/2)^4) = \frac{15N_0}{16}$. Element B, having a 40-minute half-life, completes $n = 2$ cycles in the same 80 minutes, resulting in a decayed count of $N_0(1 - (1/2)^2) = \frac{3N_0}{4}$. The ratio of decayed nuclei for A and B is:

$\frac{15N_0/16}{3N_0/4} = \frac{15}{16} \times \frac{4}{3} = \frac{5}{4}$

This ratio demonstrates that element A has decayed significantly more than element B due to the higher number of half-lives it has completed within the given time frame.

When an electron transitions from an excited state to the ground state of a hydrogen-like atom, it releases energy, causing the total energy of the system to decrease to a more negative value. According to the relationship $TE = -KE$, a decrease in total energy corresponds to an increase in the kinetic energy of the electron. Furthermore, because the potential energy is given by $PE = 2TE$, the potential energy also trends toward a more negative value, which mathematically constitutes a decrease as the electron moves closer to the nucleus.

The wavelength of radiation emitted during an electronic transition in hydrogen-like atoms is governed by the Rydberg formula, where the inverse wavelength is proportional to the square of the atomic number $Z$. For a transition from energy level $n=2$ to $n=1$, the relationship is given by:
$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$
Simplifying this expression results in $\lambda = \frac{4}{3RZ^2}$. Since Hydrogen and Deuterium both possess an atomic number of $Z=1$, their emitted wavelengths are identical, resulting in $\lambda_1 = \lambda_2 = \frac{4}{3R}$. For singly ionized Helium, which has an atomic number of $Z=2$, the wavelength becomes $\lambda_3 = \frac{4}{3R(2^2)} = \frac{4}{12R}$, and for doubly ionized Lithium with $Z=3$, the wavelength is $\lambda_4 = \frac{4}{3R(3^2)} = \frac{4}{27R}$. Comparing these values, we can see that $\lambda_1$ is equal to $\lambda_2$, is four times $\lambda_3$, and is nine times $\lambda_4$, which confirms the relationship $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

Determining the work function requires equating the incident photon energy with the sum of the metal's work function and the maximum kinetic energy of the ejected electrons. A hydrogen atom transitioning from the $n=3$ to $n=2$ energy level emits a photon with an energy of $13.6 \left( \frac{1}{4} - \frac{1}{9} \right)$ eV, which evaluates to approximately $1.89$ eV.

The maximum kinetic energy of these electrons is derived from their circular path within the magnetic field. Since the Lorentz force provides the centripetal acceleration, the relationship $mv = qBR$ holds, allowing the kinetic energy to be expressed as $KE = \frac{(qBR)^2}{2m}$. Using the provided values of $B = 3 \times 10^{-4}$ T and $R = 10.0$ mm, this calculation results in a maximum kinetic energy of $0.8$ eV. Subtracting this kinetic energy from the initial photon energy according to Einstein's photoelectric equation, $\phi = E - KE$, yields a work function of approximately $1.89 - 0.8 = 1.1$ eV.

The frequency of radiation emitted during an electronic transition is directly related to the energy difference between the initial and final states, defined by $\Delta E = h\nu$, where energy levels are proportional to $1/n^2$. For a transition between level $n$ and $n-1$, the frequency is proportional to $\left( \frac{1}{(n-1)^2} - \frac{1}{n^2} \right)$. Combining these fractions yields $\frac{n^2 - (n-1)^2}{n^2(n-1)^2}$, which simplifies to $\frac{2n-1}{n^2(n-1)^2}$. Applying the approximation $n \gg 1$, the numerator $2n-1$ effectively becomes $2n$ and the denominator $n^2(n-1)^2$ simplifies to $n^4$, resulting in a proportionality of $\frac{2n}{n^4}$, or simply $\frac{1}{n^3}$.

Einstein's mass-energy equivalence principle dictates that the energy released in a nuclear transformation is directly proportional to the mass lost during the process. Calculating the mass deficit involves finding the difference between the neutron's mass and the combined mass of the resulting proton and electron. Based on the provided values, the mass defect is $1.6725 \times 10^{-27} - (1.6725 \times 10^{-27} + 9 \times 10^{-31})$ kg, which equals $9 \times 10^{-31}$ kg. This mass is then converted into energy using the equation

$E = \Delta m c^2$

By substituting the mass deficit and the speed of light, approximately $3 \times 10^8$ m/s, into the equation, the energy release is $81 \times 10^{-15}$ Joules. Dividing this result by the conversion factor of $1.6 \times 10^{-19}$ Joules per eV gives a final energy release of approximately 0.51 MeV.

For a rotating diatomic molecule, the rotational kinetic energy is defined by the relationship $E = \frac{L^2}{2I}$, where $L$ represents the angular momentum and $I$ denotes the moment of inertia about the center of mass. The moment of inertia for this system is calculated using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$ and the separation distance $r$, such that $I = \frac{m_1 m_2}{m_1 + m_2} r^2$. By applying the quantization condition $L = nh$, we substitute this expression for angular momentum and the formula for $I$ into the energy equation to get $E = \frac{(nh)^2}{2 \left( \frac{m_1 m_2}{m_1 + m_2} r^2 \right)}$. Simplifying this expression by moving the denominator of the reduced mass term to the numerator yields the final energy as $\frac{(m_1 + m_2) n^2 h^2}{2 m_1 m_2 r^2}$.

When an electron in a hydrogen atom is excited to a higher energy level, it releases energy as it transitions back to lower energy states, producing a series of distinct spectral lines. The total number of unique transitions can be determined by the formula $\frac{n(n-1)}{2}$, where $n$ represents the principal quantum number of the excited state. For an electron transitioning from $n=4$, the calculation is:

$\frac{4(4-1)}{2} = 6$

This sequence of transitions accounts for every possible pathway from the excited shell down to the ground state, resulting in exactly six spectral lines.

The energy required to excite an electron within a hydrogen-like ion is determined by the atomic number of the nucleus and the difference between the reciprocal squares of the initial and final energy levels. For $\mathrm{Li}^{++}$, the atomic number is $Z = 3$, and the transition occurs from the first Bohr orbit, where $n_1 = 1$, to the third Bohr orbit, where $n_2 = 3$. Applying the equation $\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{eV}$, we insert these values to obtain $\Delta E = 13.6 \times (3)^2 \times \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$. This calculation simplifies to $13.6 \times 9 \times \left( 1 - \frac{1}{9} \right)$, which further reduces to $13.6 \times 9 \times \frac{8}{9}$, resulting in an excitation energy of $108.8 \text{eV}$.