Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

In a Bohr atom, the energy of an emitted photon is strictly determined by the energy difference between the initial and final states of the electron. According to the relationship $E = \frac{hc}{\lambda}$, the energy of the transition is inversely proportional to the wavelength of the emitted radiation. Because the total energy difference between state $A$ and state $B$ must equal the difference in energy between transitions $A$ to $C$ and $B$ to $C$, we can express the wavelengths using the relationship $\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}$.

Substituting the provided wavelengths of $2000 \text{Å}$ for the transition from $A$ to $C$ and $6000 \text{Å}$ for the transition from $B$ to $C$, the equation becomes $\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000}$. By determining a common denominator, this simplifies to $\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000}$, which results in $\frac{2}{6000}$ or $\frac{1}{3000}$. Consequently, the wavelength of the radiation emitted during the transition of the electron from state $A$ to state $B$ is 3000 Å.

The wavelength of a photon emitted during an electron transition is inversely proportional to the energy difference between the two orbits, meaning the longest wavelength in any spectral series corresponds to the transition between adjacent energy levels where the energy gap is smallest. For the Lyman series, this transition occurs from the second orbit to the first, resulting in the relationship $\frac{1}{\lambda_L} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$, which implies $\lambda_L = \frac{4}{3R}$.

For the Balmer series, the longest wavelength occurs with the transition from the third orbit to the second, represented by $\frac{1}{\lambda_B} = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$, which implies $\lambda_B = \frac{36}{5R}$. Comparing these two values by taking the ratio $\frac{\lambda_L}{\lambda_B}$ yields $\frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36}$, resulting in a final ratio of $\frac{5}{27}$.

Distinguishing between these reactions depends on the underlying mechanism of the transformation and the associated energy changes. Nuclear fission involves heavy nuclei breaking apart into smaller, more stable fragments, which is exactly what occurs when a neutron strikes ${}^{235}_{92}\mathrm{U}$ to produce ${}^{140}_{54}\mathrm{Xe}$, ${}^{94}_{38}\mathrm{Sr}$, and additional neutrons. In contrast, chemical reactions are driven by the rearrangement of atoms and molecules rather than changes to the nucleus, exemplified by the combustion reaction $2\,\mathrm{H}_{2} + \mathrm{O}_{2} \longrightarrow 2\,\mathrm{H}_{2}\mathrm{O}$.

Nuclear fusion occurs when lighter nuclei combine to form heavier ones, a process that is exothermic with a positive Q value when it releases energy, such as when two deuterons fuse to produce helium and a neutron according to the reaction ${}^{2}_{1}\mathrm{H} + {}^{2}_{1}\mathrm{H} \longrightarrow {}^{3}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$. Alternatively, fusion can be endothermic with a negative Q value if the process requires energy input to proceed, as is the case with the reaction ${}^{1}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \longrightarrow {}^{2}_{1}\mathrm{H} + {}^{2}_{1}\mathrm{H}$.

The radioactive decay process $P \to Q \to R$ is governed by consecutive decay kinetics, where the quantity of parent isotope $P$ follows the exponential decay function $N_P(t) = N_0 e^{-\lambda_P t}$, decreasing continuously from its initial value toward zero. The intermediate isotope $Q$ starts at zero, increases to a peak as it is generated by the decay of $P$, and subsequently decreases to zero as it transforms into the stable product $R$. Finally, the stable substance $R$ accumulates over time, starting from zero and rising asymptotically to equal the initial mass of $P$ as $t \to \infty$.

The recoil of an atom following photon emission is governed by the principle of conservation of linear momentum. Because the system starts at rest, the momentum carried by the emitted photon must be balanced by an equal and opposite recoil momentum of the hydrogen atom, which leads to the relationship $v = \frac{E}{mc}$, where $v$ is the recoil speed, $E$ is the photon energy, $m$ is the atom's mass, and $c$ is the speed of light.

Given the transition from $n=2$ to $n=1$, the energy released is 10.2 eV. Converting this energy into standard units results in $10.2 \times 1.6 \times 10^{-19}$ J. Substituting the provided mass of $1.6 \times 10^{-27}$ kg and the speed of light of $3 \times 10^8$ m/s into our formula gives $v = \frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^8}$. This calculation simplifies to a recoil speed of 3.4 m/s. Since the problem states that the recoil speed is $\frac{x}{5}$ m/s, equating $\frac{x}{5} = 3.4$ yields $x = 17$.

The nuclear radius $R$ relates to the mass number $A$ through the relation $R = R_0 A^{1/3}$, where $R_0$ is a constant. Because $R_0$ is identical for all nuclei, the ratio of the radii for any two nuclei is equal to the cube root of the ratio of their mass numbers:
$\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$
By cubing both sides of this expression, we establish a direct relationship between the ratio of the cubes of the radii and the mass numbers:
$\left( \frac{R_1}{R_2} \right)^3 = \frac{A_1}{A_2}$
Substituting the given values, where the first nucleus has a radius of $4.8$ fermi and mass number $64$, and the second has a radius of $4$ fermi, the equation becomes:
$\left( \frac{4.8}{4} \right)^3 = \frac{64}{A}$
Simplifying the division inside the parenthesis results in $1.2$, and cubing $1.2$ yields $1.728$. This simplifies the equation to $1.728 = \frac{64}{A}$, which rearranges to $A = \frac{64}{1.728}$. To express this in the form $\frac{1000}{x}$, we convert $1.728$ into the fraction $\frac{1728}{1000}$, leading to $A = \frac{64 \times 1000}{1728}$. Dividing $1728$ by $64$ gives $27$, resulting in $A = \frac{1000}{27}$. Identifying $x$ by comparison shows that $x$ is $27$.

Because the nucleus is initially at rest, the system must conserve its total linear momentum of zero throughout the disintegration process. This physical requirement necessitates that the two fragments travel in opposite directions so that their individual momenta cancel each other out, which is expressed by the conservation equation $m_1 v_1 + m_2 v_2 = 0$. Given the mass ratio of $2:1$, this relationship dictates that the ratio of the fragments' speeds must be the inverse of their mass ratio, resulting in a speed ratio of $1:2$.

In the Bohr model, the electrostatic attraction between the nucleus and the electron dictates a fixed energy relationship where the kinetic energy is exactly half the magnitude of the potential energy. This fundamental balance relies on the nature of the Coulombic force and holds true regardless of the orbital index, making the ratio independent of the specific energy level. For an electron in the fifth excited state, this invariant relationship dictates that the ratio of kinetic energy to the absolute value of potential energy is $\frac{1}{2}$.

The magnetic moment of an orbiting electron is defined by the product of its equivalent current and the area of the orbit, which simplifies to the expression $\mu = \frac{evr}{2}$. Based on Bohr's theory, the velocity of an electron is inversely proportional to the principal quantum number, while the orbital radius is proportional to the square of the principal quantum number, giving us $v \propto \frac{1}{n}$ and $r \propto n^2$. By substituting these relationships into the expression for magnetic moment, we find that $\mu$ is proportional to $\frac{1}{n} \cdot n^2$, which simplifies to $n$. Therefore, the value of the exponent $x$ is 1.

Einstein's principle of mass-energy equivalence states that any mass lost in a reaction is converted into energy according to the formula $E = \Delta mc^2$. To find the energy released for a mass defect of 0.4 g, we first convert the mass into standard SI units, which is 0.4 x 10^-3 kg. Substituting this mass along with the speed of light of 3 x 10^8 m/s gives the equation $E = 0.4 \times 10^{-3} \times (3 \times 10^8)^2$. Performing this calculation yields 3600 x 10^7 kWs. Since 1 kWh is equal to 3600 kWs, dividing the total energy by 3600 converts the value into kilowatt-hours, resulting in 1 x 10^7 kWh and identifying the missing value n as 1.

The electrostatic attraction between the nucleus and an electron provides the centripetal force necessary to sustain circular motion, expressed by the equation $\frac{Kq_1q_2}{r^2} = \frac{mv^2}{r}$. Rearranging this to isolate the product $mv^2$ gives $mv^2 = \frac{Kq_1q_2}{r}$, and because angular momentum is defined as $L = mvr$, we can square this definition to find $L^2 = m^2v^2r^2$, which is equivalent to $L^2 = m(mv^2)r^2$. By substituting the expression for $mv^2$ into this relationship, the equation becomes $L^2 = mKq_1q_2r$, confirming that the angular momentum of the electron is directly proportional to $\sqrt{r}$.

The Paschen series consists of spectral lines generated when electrons in a hydrogen atom undergo a transition to the third energy level. Because wavelength is inversely proportional to the energy of the emitted photon, the longest wavelength in this series corresponds to the smallest possible energy transition, which occurs when an electron drops from the fourth orbit to the third orbit.

Applying the Rydberg formula $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, we set $n_1 = 3$ for the Paschen series and $n_2 = 4$ for the longest wavelength. Substituting the given constant $R_H = 1.097 \times 10^7$ into the expression results in the following equation:

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$

Calculating the difference between the inverse squares, $\frac{1}{9} - \frac{1}{16}$, gives $\frac{7}{144}$. Rearranging the equation to solve for the wavelength yields $\lambda = \frac{144}{7 \times 1.097 \times 10^7}$, which results in a value of $1.876 \times 10^{-6}$ m.

Einstein's principle of mass-energy equivalence relates the binding energy of a nucleus to its mass defect through the equation $E = \Delta m c^2$. Using the binding energy $E = 18 \times 10^8$ J and the speed of light $c = 3 \times 10^8$ m/s, we determine that $c^2 = 9 \times 10^{16}$ m$^2$/s$^2$. Dividing the binding energy by this squared constant results in a mass defect of $\Delta m = 2 \times 10^{-8}$ kg. Converting this mass into micrograms by multiplying by $10^9$ yields a final difference of 20 micrograms.

The Balmer series encompasses all emission transitions where electrons fall to the $n=2$ energy level. For a hydrogen atom starting in the ground state ($n=1$) to emit radiation within this series, it must first be excited to at least the $n=3$ energy level so that it can subsequently drop down to $n=2$. The minimum energy required to initiate this process is the energy needed to transition the electron from $n=1$ to $n=3$, which is $\Delta E = 12.1\,\text{eV}$.

The wavelength of radiation emitted during an electronic transition is inversely proportional to the energy difference between the initial and final energy levels. For the Paschen series, electrons transition to the third energy shell, defined as $n_1 = 3$. To find the longest wavelength, we must identify the smallest possible energy gap, which occurs when an electron drops from the immediate next shell, $n_2 = 4$, down to $n_1 = 3$. Applying the Rydberg formula for a hydrogen atom, where $Z = 1$:

$\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$

Simplifying the terms inside the parentheses yields $\frac{1}{\lambda} = R \left( \frac{1}{9} - \frac{1}{16} \right)$, which further calculates to $\frac{1}{\lambda} = R \left( \frac{16 - 9}{144} \right)$, resulting in $\frac{1}{\lambda} = \frac{7R}{144}$. Inverting this expression to solve for the wavelength gives $\lambda = \frac{144}{7R}$, identifying the value of $\alpha$ as 144.

Nuclear fusion releases energy because the total mass of the resulting nucleus is slightly less than the sum of the masses of the reactant particles. This mass discrepancy, or mass defect, converts directly into energy according to the principle of mass-energy equivalence. The total reactant mass is $3 \times 4.002603 \ \mathrm{u} = 12.007809 \ \mathrm{u}$, so the mass defect relative to the 12 u carbon nucleus is $\Delta m = 12.007809 \ \mathrm{u} - 12 \ \mathrm{u} = 0.007809 \ \mathrm{u}$. Multiplying this defect by the conversion factor of 931 MeV/u results in an energy release of $0.007809 \times 931 \ \mathrm{MeV} = 7.27 \ \mathrm{MeV}$, which is equivalent to $727 \times 10^{-2} \ \mathrm{MeV}$.

Energy release in fission occurs because the system transitions to a state of higher stability, where the total binding energy of the products exceeds that of the parent nucleus. The net energy released is given by the expression $Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactant}}$.

The parent nucleus has a total binding energy of 236 multiplied by 7.6 MeV. Since the fission produces two fragments of 118 nucleons each, the final total binding energy is 2 multiplied by 118 multiplied by 8.6 MeV, which is equivalent to 236 multiplied by 8.6 MeV. Subtracting the initial total binding energy from the final total binding energy yields $Q = 236 \times 8.6 - 236 \times 7.6$, leading to 236 MeV.

To detect lines belonging to the Balmer series in an emission spectrum, a hydrogen atom must transition to the $n=2$ energy level from a higher orbit. This necessitates that the atom be initially excited from its ground state, $n=1$, to at least the $n=3$ energy level, as the primary transition resulting in a Balmer line occurs when an electron falls from $n=3$ to $n=2$. If the energy provided by the bombarding electrons is insufficient to promote the atom to the $n=3$ state, the atom cannot occupy the higher energy level required to initiate Balmer series emissions.

The minimum potential difference $V$ must correspond to the energy difference between the $n=1$ and $n=3$ levels. Using the energy formula for the hydrogen atom, this energy gap is given by $\Delta E = 13.6 \left(1 - \frac{1}{3^2}\right)$ eV. Calculating this value gives $V = \frac{13.6 \times 8}{9}$, which simplifies to approximately 12.09 V, or 12.1 V. Given that this potential is defined as $\frac{\alpha}{10}$ V, we equate 12.1 with $\frac{\alpha}{10}$ to determine that $\alpha = 121$.

Nuclear reactions must adhere to the conservation laws of mass number and atomic number, which dictate that the total sum of these quantities remains invariant across both sides of the reaction equation. In the given fission reaction, the reactant side shows a total mass number of 236, while the known products contribute a sum of $141 + 92 = 233$. Consequently, the three emitted particles must account for the remaining mass difference of $236 - 233 = 3$, signifying that each particle has a mass number of 1. Performing a similar check for the atomic number, the reactant side total of 92 matches the sum of the known products $56 + 36 = 92$, which implies each emitted particle possesses an atomic number of 0. Given these properties, the particles are identified as neutrons.

When a hydrogen atom transitions from an excited state to a lower one, the conservation of energy and momentum requires that the atom must recoil to compensate for the momentum of the emitted photon, which slightly reduces the energy available to the photon itself. The energy released during the $n=3$ to $n=2$ transition is calculated as $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \approx 1.9\ \mathrm{eV}$.

By applying the laws of conservation, the fractional shift in the wavelength of the emitted light can be related to the energy change by the expression:
$\frac{\Delta \lambda}{\lambda} = \frac{\Delta E}{2mc^2}$
Substituting the transition energy $\Delta E = 1.9 \times 1.6 \times 10^{-19}\ \mathrm{J}$, the mass $m = 1.6 \times 10^{-27}\ \mathrm{kg}$, and the speed of light $c \approx 3 \times 10^8\ \mathrm{m/s}$ into this formula yields a fractional change of approximately $10^{-9}$. Since the percentage change is determined by multiplying this fractional value by $100$, the result is $10^{-7}$, which identifies the value of $n$ as $7$.