Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

According to Bohr's quantization condition, the angular momentum of an electron in a stable orbit is expressed by the equation:
$mvr = \frac{nh}{2\pi}$
Since linear momentum is defined as the product of mass and velocity, we can determine its value by dividing both sides of this angular momentum relationship by the orbital radius $r$. This algebraic rearrangement yields a linear momentum of $\frac{nh}{2\pi r}$ for an electron revolving in the $n^{\text{th}}$ orbit.

In a nuclear reaction, the total energy available to the products is dictated by the combination of the incoming projectile's kinetic energy, $K_p$, and the energy released or consumed during the interaction, $Q$. Since conservation of energy and momentum must be maintained, and the kinetic energy of the projectile is always positive, the net energy of the system must remain positive for the reaction to be physically possible. This constraint forces the sum of the projectile's kinetic energy and the Q-value to be greater than zero, expressed as $K_p + Q > 0$.

When a radioactive nucleus decays through multiple independent channels simultaneously, the total decay rate is the additive sum of the individual decay constants, expressed as $\lambda_{eff} = \lambda_1 + \lambda_2$. Given the inverse relationship between the decay constant and half-life, where $\lambda = \frac{\ln 2}{T_{1/2}}$, the combined effective half-life relates to the individual half-lives through the equation $\frac{1}{T_{eff}} = \frac{1}{T_1} + \frac{1}{T_2}$. Applying the provided values of $3.0$ hours and $4.5$ hours, the calculation becomes $\frac{1}{T_{eff}} = \frac{1}{3.0} + \frac{1}{4.5}$. Finding a common denominator results in $\frac{1}{T_{eff}} = \frac{4.5 + 3.0}{3.0 \times 4.5} = \frac{7.5}{13.5}$. Inverting this expression yields an effective half-life of $\frac{13.5}{7.5}$, which equals $1.8$ hours.

The total energy of an electron in a hydrogen atom is always half its potential energy, as given by the relationship $E = \frac{PE}{2}$. Since the electron starts in the first excited state, where $n = 2$, its initial energy is $E_i = \frac{-13.6}{2^2} = -3.4$ eV. After absorbing the photon, the electron reaches a final state with potential energy $-1.08$ eV, which corresponds to a final total energy of $E_f = \frac{-1.08}{2} = -0.54$ eV. The energy of the absorbed photon is the difference between these states, calculated as $\Delta E = E_f - E_i = -0.54 - (-3.4) = 2.86$ eV, and equating this to $x \times 10^{-2}$ eV results in the equation $2.86 = x \times 10^{-2}$, which yields $x = 286$.

Radioactive decay dictates that the number of remaining nuclei decreases by a factor of $(\frac{1}{2})^n$, where $n$ represents the number of half-lives passed. Since 16 s have elapsed, substance $A$ experiences $\frac{16}{4} = 4$ half-lives, leaving $(\frac{1}{2})^4 = \frac{1}{16}$ of its initial atoms, whereas substance $B$ experiences $\frac{16}{8} = 2$ half-lives, leaving $(\frac{1}{2})^2 = \frac{1}{4}$ of its initial atoms. Because both samples start with an equal mass of $10^{-2}$ kg, the initial number of atoms is inversely proportional to their atomic weights. Given the ratio of atomic weights of $A$ to $B$ is $\frac{1}{2}$, substance $A$ initially contains twice as many atoms as $B$. Putting these together, the ratio of remaining atoms $\frac{N_A}{N_B}$ is calculated as $(\frac{N_{0A}}{16}) / (\frac{N_{0B}}{4})$, which simplifies to $2 \times \frac{4}{16} = \frac{1}{2}$. To find the final mass ratio, we multiply this atom ratio by the atomic weight ratio, resulting in $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. As this equals $\frac{25}{100}$, the value of $x$ is 25.

Radioactive decay is governed by an exponential relationship where the activity of a sample decreases over time at a rate proportional to its current amount, represented by the formula $A = A_0 e^{-\lambda t}$. In this scenario, $A$ is 2250 disintegrations per minute, $A_0$ is 4250 disintegrations per minute, and the time $t$ is 10 minutes.

With these parameters, the equation $2250 = 4250 e^{-10\lambda}$ rearranges to $e^{10\lambda} = \frac{4250}{2250}$, which is approximately $1.888$. Taking the natural logarithm on both sides results in
$10\lambda = \ln(1.888)$
Converting the natural logarithm to base-10 using the standard conversion factor of $2.303$ and the provided value $\log_{10} 1.88 = 0.274$, the equation becomes $10\lambda = 2.303 \times 0.274$, yielding a value of approximately $0.631$. Dividing this result by the 10 minutes elapsed gives a final decay constant of approximately $0.063\ \mathrm{min}^{-1}$.

Radioactive decay is governed by the principle that a substance reduces to half its remaining amount after every specific period defined as the half-life. Since the sample has decayed by $\frac{7}{8}$ of its original quantity, the remaining amount is $\frac{1}{8}$ of the starting value. This remaining fraction represents three successive halving events, as $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$, meaning three full half-lives have passed within the 15-minute timeframe. Dividing the total time elapsed by the number of half-lives, $T_{1/2} = \frac{15}{3}$, results in a half-life duration of 5 minutes.

According to the Bohr model, the energy levels of a hydrogen atom are determined by the principal quantum number $n$ through the relationship $E_n = -\frac{13.6}{n^2} \text{ eV}$. In the ground state where $n=1$, the atom occupies an energy level of $-13.6 \text{ eV}$. When the atom absorbs $12.09 \text{ eV}$ of energy, the electron transitions to a higher energy level, resulting in a final energy of $E_f = -13.6 \text{ eV} + 12.09 \text{ eV} = -1.51 \text{ eV}$. Setting this equal to the energy formula $-\frac{13.6}{n^2} = -1.51$ yields $n^2 = 9$, which identifies the new state as $n=3$.

Bohr's quantization condition states that the orbital angular momentum $L$ is given by $L = \frac{nh}{2\pi}$. Therefore, the increase in angular momentum when moving from the $n=1$ state to the $n=3$ state is the difference between these two quantized values:

$\Delta L = \frac{3h}{2\pi} - \frac{1h}{2\pi} = \frac{h}{\pi}$

Using Planck's constant $h = 6.63 \times 10^{-34} \text{ J s}$ and taking $\pi \approx 3.14$, the change in angular momentum is calculated as $\frac{6.63 \times 10^{-34}}{3.14}$, which equals $2.11 \times 10^{-34} \text{ J s}$.

Radioactive decay follows a predictable exponential path where the activity of a sample decreases over time. The fundamental relationship is defined by the expression $A = A_0 e^{-\lambda t}$, where $A$ is the final activity, $A_0$ is the initial activity, $\lambda$ is the decay constant, and $t$ is the time elapsed. The decay constant itself is determined by the half-life as $\lambda = \frac{\ln 2}{T_{1/2}}$.

With an initial activity of $6.4 \times 10^{-4}$ curie and a half-life of 5 days, we need to find the duration required for the activity to reach $5 \times 10^{-6}$ curie. Setting up the ratio of the final activity to the initial activity gives $\frac{5 \times 10^{-6}}{6.4 \times 10^{-4}} = e^{-\lambda t}$, which simplifies to $0.0078125 = e^{-\lambda t}$. By taking the natural logarithm on both sides, we arrive at $\ln(0.0078125) = -\lambda t$. Substituting the decay constant value derived from the half-life, we solve for $t$ to find that the total time required is 35 days.

The radius of an atomic nucleus relates to its mass number $A$ through the fundamental expression $R = R_0 A^{1/3}$, where $R_0$ is the nuclear radius constant. Dividing both sides by $R_0$ and taking the natural logarithm of the equation yields $\ln\!\left(\frac{R}{R_0}\right) = \frac{1}{3} \ln(A)$. This mathematical form corresponds to a linear relationship $y = mx$ with a slope of $1/3$, implying the graph represents a straight line passing through the origin.

Radioactive decay functions as a purely random and spontaneous nuclear phenomenon that remains entirely independent of any external physical or chemical conditions, which invalidates the first statement. The population of undecayed nuclei follows the exponential law $N = N_0 e^{-\lambda t}$, confirming that the decay process is exponential over time. By taking the natural logarithm of this equation, we get $\ln N = -\lambda t + \ln N_0$, where the slope is $-\lambda$; since the mean lifetime is defined as $\tau = 1/\lambda$, the slope magnitude represents the reciprocal of the mean lifetime. Finally, the relationship $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ shows that the product of the decay constant and half-life is exactly $\ln 2$, which is a constant, proving the last statement incorrect.

Radioactive decay follows a predictable geometric pattern where the amount of substance is halved every half-life period. If 7/8 of the original mass has disintegrated, it implies that the remaining fraction is $1 - \frac{7}{8} = \frac{1}{8}$ of the initial amount. To determine how many half-lives have passed, we observe the successive halving of the sample: after one half-life, $\frac{1}{2}$ remains, after two, $\frac{1}{4}$ remains, and after three, $\frac{1}{8}$ remains. Consequently, since it takes three half-lives to reach this state, the total time elapsed is $3 \times 60 = 180$ days.

In nuclear fusion, energy is liberated when lighter nuclei merge to form a more stable, heavier nucleus. This energy release occurs because the product nucleus has a higher binding energy per nucleon than the original reactant nuclei, signifying that the nucleons have transitioned into a more tightly bound, lower-energy state. The total binding energy of a nucleus is obtained by multiplying the binding energy per nucleon by the total number of nucleons in that nucleus.

For the reaction where two ${}_1^2 X$ nuclei combine into one ${}_2^4 Y$ nucleus, the energy balance is determined by the total binding energy of the system. The reactants consist of two nuclei, each with 2 nucleons and a binding energy of 1.1 MeV per nucleon, resulting in a total reactant binding energy of 4.4 MeV. The product is a single nucleus with 4 nucleons, each having a binding energy of 7.6 MeV, leading to a total product binding energy of 30.4 MeV. The net energy released corresponds to the difference in binding energy between the final product and the initial reactants:

$\Delta E = 30.4 - 4.4 = 26$

Thus, the energy released in this process is 26 MeV.

The energy emitted by a hydrogen atom during an electron transition is determined by the difference in energy levels, calculated using $E_n = \frac{-13.6}{n^2}$ eV, where the photon energy is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. For the transition from the second level to the first, the emitted energy is $13.6 \left( 1 - \frac{1}{4} \right)$, which simplifies to $13.6 \times \frac{3}{4}$ eV. A transition from the highest permitted energy level, which is infinity, to the first level releases energy equal to $13.6 \left( 1 - 0 \right)$, or simply $13.6$ eV. Comparing these two results, the ratio of the photon energies is $3:4$.

The radioactive decay process follows the relationship $\frac{A}{A_0} = \left(\frac{1}{2}\right)^n$, where $A$ is the final activity, $A_0$ is the initial activity, and $n$ is the number of half-lives that have elapsed. Calculating the ratio of the final activity 2 × 10⁻⁵ Ci to the initial activity 2.56 × 10⁻³ Ci gives 1/128. Because this ratio represents the decay factor $(\frac{1}{2})^7$, we know that exactly 7 half-lives have passed. Multiplying this number by the half-life period of 5 days, the total time elapsed is $7 \times 5 = 35$ days.

Radioactive decay occurs in a predictable cycle where the quantity of a substance reduces by half after every half-life interval. Since the activity dropped to $\frac{1}{16}$ of its initial value, we can relate this reduction to the number of half-lives $n$ using the equation $\left(\frac{1}{2}\right)^n = \frac{1}{16}$. Solving for $n$ reveals that 4 half-lives have elapsed during the total time of 30 years. Dividing this total duration by the number of elapsed half-lives results in a half-life period of $\frac{30}{4}$, which is 7.5 years.

The speed of an electron in a hydrogen-like orbit is governed by the relationship $v \propto \frac{Z}{n}$, where $Z$ is the atomic number and $n$ is the principal quantum number. For the third orbit of $\mathrm{He}^{+}$, where $Z=2$ and $n=3$, compared to the third orbit of a hydrogen atom, where $Z=1$ and $n=3$, we see that the principal quantum number is constant in both systems. Consequently, the speed ratio is determined solely by the ratio of their respective atomic numbers, resulting in a value of 2:1.

In nuclear fission, energy is released when a heavy nucleus splits into fragments that are more tightly bound, as reflected by an increase in the binding energy per nucleon. This gain in stability is liberated as energy, which is determined by calculating the difference between the total binding energy of the resulting fragments and the initial parent nucleus.

The total binding energy of the parent nucleus is obtained by multiplying its mass number of 220 by the given binding energy of 5.6 MeV, resulting in $220 \times 5.6$ MeV. Similarly, the total binding energy of the fragments is found by summing their individual energies, calculated as $(105 \times 6.4 + 115 \times 6.4)$ MeV. The energy released, Q, corresponds to the difference between these two totals, expressed as:

$Q = (105 \times 6.4 + 115 \times 6.4) - (220 \times 5.6)\,\text{MeV}$

Performing this calculation yields a final result of 176 MeV.

Nuclear density is a fundamental constant for all atomic nuclei and does not depend on the mass number. Since the total mass of a nucleus is proportional to its mass number $A$ and its volume is proportional to the cube of the radius, where $R = R_0 A^{1/3}$, the volume itself scales linearly with $A$. Calculating the density $\rho$ by dividing the total mass by the volume results in the following expression:

$\rho = \frac{m A}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{3m}{4 \pi R_0^3}$

Because the mass number $A$ appears in both the numerator and the denominator, it cancels out entirely, demonstrating that nuclear density is independent of the mass number. Therefore, the ratio of the densities for any two nuclei is $1:1$.

Radioactive transformations follow specific conservation laws: an alpha particle emission reduces the mass number by four and the atomic number by two, whereas a beta-minus decay increases the atomic number by one without altering the mass number, and gamma decay leaves both values unchanged. Starting with the nucleus $D$ defined by mass number $182$ and atomic number $74$, we track the total change across two alpha decays, one beta-minus decay, and one gamma emission. The cumulative effect on the mass number is calculated as $182 - (2 \times 4) = 174$, while the final atomic number is determined by $74 - (2 \times 2) + 1 = 71$.