Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Radioactive decay is governed by the law $N = N_0 e^{-\lambda t}$, which describes how the number of nuclei decreases exponentially over time. Since both substances A and B start with the same initial population $N_0$, the ratio of their remaining nuclei is $\frac{N_A}{N_B} = \frac{e^{-5\lambda t}}{e^{-\lambda t}}$, which simplifies to $e^{-4\lambda t}$. Equating this result to the target value of $\frac{1}{e^2}$ gives the expression $e^{-4\lambda t} = e^{-2}$, and solving the resulting equality $-4\lambda t = -2$ yields the time $t = \frac{1}{2\lambda}$.

Nuclear stability is determined by the binding energy per nucleon, where a higher value signifies a more stable, lower-energy state. To assess the energy change during the fission of $\mathrm{Ne}^{20} \rightarrow 2\mathrm{He}^{4} + \mathrm{C}^{12}$, we compare the total binding energies of the reactant and the products. The total binding energy of the reactant is $20 \times 8.03 = 160.6 \text{ MeV}$, while the total binding energy of the products is calculated as follows:

$(2 \times 4 \times 7.07) + (12 \times 7.86) = 150.88 \text{ MeV}$

The difference between the reactant and product binding energies, $160.6 - 150.88 = 9.72 \text{ MeV}$, indicates that the reactant has a higher binding energy and is therefore more stable than the product nuclei. Consequently, the reaction is endothermic and requires an input of external energy to proceed.

Radioactive decay is characterized by a constant activity reduction where the count rate halves over every specific duration known as the half-life. Given an initial count rate of 1600 counts per second that diminishes to 100 counts per second over an 8-second interval, we can identify the decay factor as $\frac{100}{1600} = \frac{1}{16}$. Because $\frac{1}{16}$ is equivalent to $\left(\frac{1}{2}\right)^4$, it follows that exactly 4 half-lives occurred during these 8 seconds. This confirms that the half-life of the radioactive source is 2 seconds, calculated by dividing the 8-second interval by the 4 elapsed half-lives.

To determine the count rate at 6 seconds, we find how many half-lives have passed by dividing the time of 6 seconds by the 2-second half-life, yielding 3 half-lives. We then apply the decay equation $A = A_0 \left(\frac{1}{2}\right)^n$, where $n = 3$. Substituting the initial value of 1600 into the expression $1600 \times \left(\frac{1}{2}\right)^3$ results in $1600 \times \frac{1}{8}$, which gives an activity of 200 counts per second.

Radioactive decay is a statistical process where the proportion of nuclei remaining in a sample decreases by half during each half-life interval, meaning the fraction of nuclei remaining is $N = N_0 (\frac{1}{2})^n$, where $n$ is the number of half-lives elapsed. The quantity of decayed nuclei is defined as the initial population minus the remaining population, represented by $N_{dec} = N_0 - N = N_0 (1 - (\frac{1}{2})^n)$.

For nucleus A, which has a half-life of 10 minutes, the number of half-lives elapsed in 60 minutes is $n_A = \frac{60}{10} = 6$. Consequently, the decayed fraction is $1 - (\frac{1}{2})^6 = 1 - \frac{1}{64} = \frac{63}{64}$. For nucleus B, which has a half-life of 20 minutes, the number of half-lives elapsed in 60 minutes is $n_B = \frac{60}{20} = 3$. The decayed fraction for B is $1 - (\frac{1}{2})^3 = 1 - \frac{1}{8} = \frac{7}{8}$.

To find the ratio of decayed nuclei for A and B, we divide their respective decayed fractions:
$\frac{N_{dec, A}}{N_{dec, B}} = \frac{63/64}{7/8} = \frac{63}{64} \times \frac{8}{7} = \frac{9}{8}$

The activity of a radioactive sample at any given moment is described by the equation $R = R_0 e^{-\lambda t}$, where $R_0$ represents the initial activity and $\lambda$ denotes the decay constant. When comparing two radioactive substances, A and B, that start with equal activities, the ratio of their activities at time $t$ simplifies to $\frac{R_B}{R_A} = \frac{R_0 e^{-\lambda_B t}}{R_0 e^{-\lambda_A t}}$, which reduces further to $e^{-(\lambda_B - \lambda_A)t}$. Since the problem specifies that this ratio decays as $e^{-3t}$, we can directly equate the exponents to find that the difference between the decay constants is $\lambda_B - \lambda_A = 3$.

To determine the half-life of substance B, we first find the decay constant of substance A. Using the standard relationship $T_{1/2} = \frac{\ln 2}{\lambda}$ and given that the half-life of A is $\ln 2$, we calculate $\lambda_A = \frac{\ln 2}{\ln 2} = 1$. Substituting this value back into the expression $\lambda_B - \lambda_A = 3$, we find that $\lambda_B = 3 + 1 = 4$. Consequently, the half-life of substance B is calculated as $T_{1/2, B} = \frac{\ln 2}{\lambda_B}$, which results in $\frac{\ln 2}{4}$.

Bohr's quantization condition requires that the angular momentum of an orbiting particle must be an integer multiple of $\frac{h}{2\pi}$. For a central potential $U(r) = \frac{1}{2} k r^2$, the force magnitude $F = kr$ equates to the centripetal force $\frac{mv^2}{r}$, yielding the relation $mv^2 = kr^2$. Combining these expressions by substituting $v = \frac{nh}{2\pi mr}$ into the force equation results in $r^2 \propto n$, which simplifies to $r_n \propto \sqrt{n}$. The total energy is the sum of kinetic and potential energy, $E = \frac{1}{2} mv^2 + \frac{1}{2} k r^2$, and substituting $mv^2 = kr^2$ shows that $E = kr^2$. Since $r^2$ is proportional to $n$, it follows that the energy levels scale as $E_n \propto n$.

The principle of energy conservation dictates that for an ion to absorb a photon, the photon's energy must precisely equal the difference in energy levels of the receiving ion. For a hydrogen atom transitioning from $n=2 \to n=1$, the energy of the emitted radiation is calculated using the formula $E = 13.6 \times Z^2 \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, which yields 10.2 eV. To determine which helium ion transition absorbs this exact amount of energy, we apply the same Rydberg formula with $Z=2$ for the $He^+$ ion. Testing the transition from $n=2 \to n=4$ gives an energy value of $13.6 \times 2^2 \times \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \times 4 \times \left( \frac{3}{16} \right) = 10.2$ eV, confirming that this specific jump accounts for the incident radiation.

The energy of an electron in a hydrogen-like ion is determined by the formula $E_n = -13.6 \frac{Z^2}{n^2}$ eV, where $Z$ is the atomic number and $n$ is the principal quantum number. For a $He^+$ ion, the atomic number $Z$ is 2, and since the ion is in its first excited state, the principal quantum number is $n = 2$. Substituting these values into the equation gives $E_2 = -13.6 \frac{2^2}{2^2}$ eV, resulting in a total energy of -13.6 eV. The ionization energy is the amount of energy required to remove the electron from this state, which is equal to the magnitude of the total energy, or 13.6 eV.

The Rydberg formula defines the relationship between the wavelength of spectral lines and the energy states of an electron, specifically given by $\frac{1}{\lambda} = R\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. For the first Balmer line transition from $n=3$ to $n=2$, this leads to $\frac{1}{\lambda_1} = R\left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$. By using the given wavelength of 660 nm in this equation, the reciprocal of the Rydberg constant can be expressed as $\frac{1}{R} = \frac{5 \times 660}{36}$ nm.

The second Balmer line corresponds to a transition from $n=4$ to $n=2$, resulting in the equation $\frac{1}{\lambda_2} = R\left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R}{16}$. Isolating the wavelength yields $\lambda_2 = \frac{16}{3R}$. Substituting the previously derived value for the reciprocal of the Rydberg constant into this expression gives $\lambda_2 = \frac{16}{3} \times \frac{5 \times 660}{36}$, which calculates to 488.9 nm.

Radioactive decay follows the exponential law $N(t) = N_0 e^{-\lambda t}$, where the quantity of remaining nuclei is determined by the decay constant and elapsed time. Given that both materials start with the same initial number of nuclei $N_0$, the ratio of the remaining nuclei of material A to material B is found by dividing their individual decay expressions: $\frac{N_A}{N_B} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-9\lambda t}$. Setting this ratio equal to $\frac{1}{e}$, which is equivalent to $e^{-1}$, allows us to equate the exponents such that $-9\lambda t = -1$. Solving this equation for the elapsed time yields $t = \frac{1}{9\lambda}$.

Radioactive decay processes strictly obey the laws of conservation of mass number and atomic number. Every alpha decay event removes 4 units from the mass number and 2 units from the atomic number, whereas every beta decay event leaves the mass number unaffected while increasing the atomic number by 1, effectively transforming a neutron into a proton.

Starting with the initial nucleus ${}^{232}_{90}\mathrm{Th}$, the emission of 6 alpha particles reduces the mass number by $6 \times 4 = 24$, resulting in a final mass number of $232 - 24 = 208$. Simultaneously, the atomic number decreases by $6 \times 2 = 12$ due to these alpha emissions and then increases by 4 due to the 4 beta particles, resulting in a final atomic number of $90 - 12 + 4 = 82$. These calculations identify the final nucleus as having a mass number of 208 and an atomic number of 82.

The total number of spectral lines produced during de-excitation is determined by the transitions from an excited state $n$ down to the ground state, governed by the formula $\frac{n(n-1)}{2}$. Given that six lines are observed, we solve $\frac{n(n-1)}{2} = 6$ to find that the electron was excited to the $n=4$ level. For a hydrogen-like ion such as $\mathrm{Li}^{++}$ with $Z=3$, the wavelength $\lambda$ of the radiation required to excite the electron from $n=1$ to $n=4$ is determined by the Rydberg relation, where $\lambda = \frac{914\,\text{\AA}}{Z^2} \times \left( \frac{1}{1^2} - \frac{1}{4^2} \right)^{-1}$. Substituting $Z=3$ into this expression yields $\lambda = \frac{914\,\text{\AA}}{9} \times \frac{16}{15}$, resulting in a wavelength of approximately $108.3\,\text{\AA}$, which is equivalent to $10.8\,\mathrm{nm}$.

Absorbing a photon with a specific wavelength causes an electron in a hydrogen atom to jump to a higher energy level, where the orbital radius is directly governed by the square of the principal quantum number. Calculating the energy of the incident photon using the expression $E = \frac{hc}{\lambda}$ results in $E = \frac{12500}{980} = 12.75\text{ eV}$. Adding this energy to the ground state energy of $-13.6\text{ eV}$ shifts the electron to an energy level of $-0.85\text{ eV}$. Substituting this into the Bohr energy relation $E_n = \frac{-13.6}{n^2}\text{ eV}$ identifies the destination as the $n = 4$ state. Since the radius of an orbit in a hydrogen atom follows the relationship $r_n = n^2 a_0$, the resulting radius is $16a_0$.

Electronic transitions in an ion involve the emission of photons that carry an energy equivalent to the difference between two specific energy levels. For a helium ion, the energy of a state with quantum number $n$ is determined by the formula $E_n = \frac{-4 \times 13.612}{n^2}$ eV. Converting the first emitted photon's wavelength of 108.5 nm into energy using the relation $E = \frac{1240}{\lambda}$, we obtain a value of approximately 11.43 eV. This energy change corresponds to an electron transition from $n = 5$ down to $n = 2$, where the energy difference between these two levels is consistent with the calculated photon energy. Since the second photon with a wavelength of 30.4 nm matches the energy required for the subsequent transition from $n = 2$ to the ground state of $n = 1$, the initial excited state must have been $n = 5$.

Bohr's quantization condition requires that the circumference of an electron's orbit must be an integer multiple of its de Broglie wavelength, which is represented by the formula $n \lambda = 2 \pi r_n$. In the second excited state of a hydrogen atom, the principal quantum number is $n = 3$. Given that the radius is $4.65$ \AA, we can set up the equation $3 \lambda = 2 \pi \times 4.65$. Solving this calculation yields a de Broglie wavelength of approximately $9.7$ \AA.

The wavelength of radiation emitted during an electronic transition in a hydrogen-like atom is governed by the Rydberg formula, where the inverse of the wavelength is proportional to the difference between the inverse squares of the final and initial principal quantum numbers. For an electron transitioning from the M shell, corresponding to $n=3$, to the L shell, corresponding to $n=2$, the reciprocal wavelength is expressed as:
$\frac{1}{\lambda} = K \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) = K \left( \frac{1}{4} - \frac{1}{9} \right) = K \left( \frac{5}{36} \right)$
When the electron transitions from the N shell, where $n=4$, to the L shell, the new wavelength $\lambda'$ is determined by:
$\frac{1}{\lambda'} = K \left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = K \left( \frac{1}{4} - \frac{1}{16} \right) = K \left( \frac{3}{16} \right)$
By relating these two expressions, the ratio of the reciprocals is $\frac{1/\lambda'}{1/\lambda} = \frac{K(3/16)}{K(5/36)}$, which simplifies to $\frac{\lambda}{\lambda'} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20}$. Rearranging this ratio shows that $\lambda' = \frac{20}{27} \lambda$.

The de Broglie wavelength $\lambda_n$ of an electron in the $n$-th state is proportional to $n$, which implies that the term $1/n^2$ is proportional to $1/\lambda_n^2$. The wavelength of the emitted photon $\text{wedge}_n$ resulting from a transition to the ground state follows the Rydberg relation:
$\frac{1}{\text{wedge}_n} = R\left(1 - \frac{1}{n^2}\right)$
Rearranging this as $\text{wedge}_n = \frac{1}{R}(1 - \frac{1}{n^2})^{-1}$ and applying the binomial approximation for large $n$ yields $\text{wedge}_n \approx \frac{1}{R}(1 + \frac{1}{n^2})$. Substituting the relationship between $n$ and $\lambda_n$ into this expression results in $\text{wedge}_n \approx A + \frac{B}{\lambda_n^2}$, where $A$ and $B$ are constants.

In the Bohr model, the radius of the $n^{th}$ orbit is given by $r_n \propto \frac{1}{m}$, so $m_\mu = 200 m_e$ leads to $r_\mu = \frac{r_e}{200}$, confirming statement (a). The orbital velocity $v_n$ is independent of the particle's mass, so the muon's speed is equal to the electron's, making statement (b) incorrect.

The ionization energy is defined by:
$E_n = \frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$
Since $E_n \propto m$, the ionization energy of the muonic atom is 200 times that of the hydrogen atom, confirming statement (c). With momentum $p = mv$ and invariant orbital speed $v$, the momentum scales linearly with mass, meaning $p_\mu = 200 p_e$, which confirms statement (d).

The principle of conservation of linear momentum requires that the total momentum of the system remains zero, meaning the magnitudes of momentum for the two product nuclei must be equal, represented by $m_1 v_1 = m_2 v_2$. This implies a mass ratio of $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{27}{8}$. Given that nuclear mass is proportional to the cube of the radius ($m \propto R^3$), the relationship between the radii is governed by:
$\left(\frac{R_1}{R_2}\right)^3 = \frac{m_1}{m_2} = \frac{27}{8}$
Taking the cube root of both sides leads to the radius ratio of $\frac{R_1}{R_2} = \frac{3}{2}$.

The relationship between the activity $A$ of a radioactive sample, the number of nuclei $N$, and the decay constant $\lambda$ is given by $A = \lambda N$. Since the decay constant is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}}$, the activity can be expressed as:
$A = \frac{N \ln 2}{T_{1/2}} \implies T_{1/2} = \frac{N \ln 2}{A}$
For the two samples $S_1$ and $S_2$, the ratio of their half-lives is determined by comparing their respective nuclear counts and activities:
$\frac{T_1}{T_2} = \frac{N_1}{N_2} \times \frac{A_2}{A_1}$
Substituting the given values $N_1 = 2N_2$, $A_1 = 5\,\mu\text{Ci}$, and $A_2 = 10\,\mu\text{Ci}$ yields:
$\frac{T_1}{T_2} = \left(\frac{2N_2}{N_2}\right) \times \left(\frac{10}{5}\right) = 2 \times 2 = 4$
This indicates that the half-life of $S_1$ is four times that of $S_2$, which is satisfied by the values 20 years and 5 years, respectively.