Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Atoms of A decay into atoms of B, while atoms of B simultaneously decay into atoms of C, making the population of B a dynamic balance between production and loss. The rate of change in the number of atoms of B is defined by the production rate from A minus its own decay rate, expressed as $\frac{dN_B}{dt} = \lambda N_A - \lambda N_B$. Given that the population of A at any time follows $N_A(t) = N_A(0) e^{-\lambda t}$ and the specific initial condition $N_A(0) = 2 N_B(0)$, the differential equation simplifies to $\frac{dN_B}{dt} = 2 \lambda N_B(0) e^{-\lambda t} - \lambda N_B$.

To solve for the population of B over time, rearrange the equation into $\frac{dN_B}{dt} + \lambda N_B = 2 \lambda N_B(0) e^{-\lambda t}$ and apply the integrating factor $e^{\lambda t}$. Multiplying both sides by this factor yields the derivative form $\frac{d}{dt}(N_B e^{\lambda t}) = 2 \lambda N_B(0)$. Integrating both sides with respect to time leads to $N_B e^{\lambda t} = 2 \lambda N_B(0) t + C$. Applying the initial condition at $t = 0$ reveals that $C = N_B(0)$, which results in the time-dependent population expression $N_B(t) = N_B(0) (1 + 2 \lambda t) e^{-\lambda t}$.

Dividing by the initial number of B atoms gives the evolution of the ratio as $\frac{N_B(t)}{N_B(0)} = (1 + 2 \lambda t) e^{-\lambda t}$. This function begins at a normalized value of 1 when $t = 0$ and initially rises as the relatively rapid decay of A feeds the pool of B. By setting the derivative of this function to zero, the peak population is found to occur at $t = \frac{1}{2\lambda}$. After this point, the rate of decay of B exceeds the rate of production, causing the population to decline toward zero as time progresses.

In a radioactive decay sequence where nucleus $A$ transforms into nucleus $B$, which subsequently decays into a stable nucleus $C$, the population of $B$ atoms follows a characteristic transient pattern. Because the sample initially contains zero atoms of $B$ at $t = 0$, the number of $B$ atoms begins at zero and increases as it is produced by the decay of $A$. Eventually, this accumulation reaches a maximum value before declining as the production rate from $A$ wanes and the decay of $B$ into $C$ dominates, resulting in a curve that rises and then exponentially falls back toward zero.

The number of unique spectral lines produced as electrons transition from an excited energy level to lower states is determined by the total possible combinations of transitions between those levels. We calculate this quantity using the formula:
$N = \frac{n(n-1)}{2}$
Substituting the given principal quantum number $n=6$ into this equation results in the expression $\frac{6(6-1)}{2}$. Solving this gives $\frac{6 \times 5}{2}$, which equals $15$ different wavelengths, meaning $x = 15$.

Radioactive decay is characterized by the exponential reduction of the parent nuclei, governed by the relationship $N = N_0 e^{-\lambda t}$. By defining the fraction of decayed nuclei $f$ as the ratio of the number of decayed nuclei to the initial amount, we obtain the expression $f = \frac{N_0 - N}{N_0} = 1 - e^{-\lambda t}$. To find the rate at which this fraction changes, we differentiate $f$ with respect to time, which results in $\frac{df}{dt} = \lambda e^{-\lambda t}$.

The mass of the ${}_{13}^{27}\mathrm{Al}$ nucleus is lower than the sum of its constituent protons and neutrons, a difference known as the mass defect which accounts for the energy required to disassemble the nucleus. Determining this value begins by calculating the combined mass of the 13 protons and 14 neutrons, which is $(13 \times 1.00726 + 14 \times 1.00866) \mathrm{u} = 27.21562 \mathrm{u}$. Subtracting the actual nuclear mass of $27.18846 \mathrm{u}$ from this theoretical total reveals a mass defect of $0.02716 \mathrm{u}$. Applying the energy conversion results in $27.16 \times 10^{-3} \mathrm{J}$, and rounding this figure to the nearest integer yields 27.

When an electron in an excited hydrogen atom transitions to the energy level $n=2$, it emits light that forms the Balmer series, which is the specific group of spectral lines located within the visible region. Other series, such as the Lyman series resulting from transitions to $n=1$, fall within the ultraviolet region, whereas the Paschen and Brackett series, resulting from transitions to higher energy levels, lie within the infrared region.

The radioactive decay rate, known as activity, depends on the number of available nuclei and the decay constant, which is expressed by the relation $A = \lambda N$. To find the decay constant $\lambda$, divide the natural logarithm of 2 by the half-life. Converting the half-life of 3 days into seconds gives 259,200 seconds, so the constant is $\lambda = \frac{\ln 2}{259200} \approx 2.67 \times 10^{-6}$ per second.

To determine the number of atoms $N$ in the 2 mg sample, use the molar mass of 198 g/mol and Avogadro's constant of $6 \times 10^{23}$ atoms/mol. The total number of atoms is calculated as $N = \frac{2 \times 10^{-3}}{198} \times 6 \times 10^{23} = 6.06 \times 10^{15}$. By applying the formula $A = \lambda N$ with the derived decay constant and the number of atoms, the activity of the sample is $A = 16.18 \times 10^{12}$ disintegrations per second.

When an electron orbits a nucleus, its continuous motion acts as a closed current loop, where the current represents the amount of charge flowing past a specific point per unit of time. The time period $T$ for one revolution is determined by the circumference of the orbit, $2 \pi r$, divided by the speed $v$ of the electron. By substituting this period into the fundamental definition of current, we obtain the expression $I = \frac{e}{T} = \frac{e v}{2 \pi r}$.

To find the numerical value, we substitute the electronic charge $1.6 \times 10^{-19} \text{ C}$, the speed $2.2 \times 10^{6} \text{ m/s}$, and the radius converted to meters, $0.5 \times 10^{-10} \text{ m}$, using $\frac{22}{7}$ for $\pi$. The equation becomes $I = \frac{1.6 \times 10^{-19} \times 2.2 \times 10^{6}}{2 \times \frac{22}{7} \times 0.5 \times 10^{-10}}$. After performing the arithmetic, the result is $1.12 \text{ mA}$, which translates to $112 × 10^{-2} \text{ mA}$.

In the Bohr model, the velocity of an electron in the $n$th orbit is governed by the expression $v_{n} = \frac{Z e^{2}}{2 \varepsilon_{0} n h}$, where $Z$ is the atomic number, $e$ is the charge of the electron, $h$ is Planck's constant, and $\varepsilon_{0}$ is the permittivity of free space. Given that $Z$, $e$, $h$, and $\varepsilon_{0}$ are all physical constants, the velocity $v_{n}$ varies exclusively with the principal quantum number $n$ located in the denominator, confirming that $v_{n} \propto \frac{1}{n}$.

The radioactivity of a sample is fundamentally defined by the product of the decay constant and the number of radioactive nuclei present. To determine this activity, we first calculate the total number of nuclei $N$ in the sample using the mass $m$, the atomic weight $A$, and Avogadro's number $N_{A}$ through the expression $N = \frac{m}{A} N_{A}$. With $1.50\,\text{mg}$ of $\mathrm{Au}^{198}$, this calculation provides the total count of radioactive atoms available for decay.

Next, we identify the decay constant $\lambda$, which relates to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$. Because the half-life is given as $2.7$ days, we must convert this time into seconds by calculating $2.7 \times 24 \times 3600\,\text{s}$ to maintain consistent units. The activity $A_{act}$ is then determined by:
$A_{act} = \left( \frac{\ln 2}{T_{1/2}} \right) \left( \frac{m}{A} N_{A} \right)$
Substituting the values $m = 1.5 \times 10^{-3}\,\text{g}$, $A = 198\,\text{g/mol}$, $N_{A} = 6 \times 10^{23}\,\text{mol}^{-1}$, and the converted half-life yields the decay rate in becquerels. Finally, converting this result to curies by dividing by the conversion factor $3.7 \times 10^{10}\,\text{Bq/Ci}$ leads to a value of approximately $357\,\text{Ci}$.

Radioactive decay is governed by the equation $R = R_0 e^{-\lambda t}$, and taking the natural logarithm of both sides provides the linear relationship $\ln R = \ln R_0 - \lambda t$. This equation reveals that the graph of $\ln R$ versus time yields a straight line where the slope is equal to $-\lambda$. By identifying the slope from the provided plot, we determine the decay constant to be $\lambda = \frac{3}{20} \text{ s}^{-1}$. Using the standard definition of half-life, $t_{1/2} = \frac{\ln 2}{\lambda}$, we substitute 0.693 for $\ln 2$ and 0.15 for $\lambda$ to find that the half-life is approximately 4.62 sec.

When a nucleus initially at rest undergoes alpha decay, the principles of conservation of linear momentum and energy dictate how the decay energy is partitioned. Because the parent nucleus is stationary, the alpha particle and the resulting daughter nucleus must recoil with equal and opposite momenta, ensuring the system momentum remains zero. Given that kinetic energy relates to momentum through the expression $K = \frac{p^2}{2m}$, the energy shared between the two products is inversely proportional to their respective masses.

The kinetic energy of the alpha particle is a specific fraction of the total $Q$ value, defined by the relationship $K_{\alpha} = \left( \frac{m_{daughter}}{m_{daughter} + m_{\alpha}} \right) Q$. With a parent mass number of 184 and an emitted alpha particle having a mass number of 4, the daughter nucleus retains the remaining mass number of 180. Substituting these values into the relationship yields the calculation:

$K_{\alpha} = \frac{180}{180 + 4} \times 5.5$

Solving this expression confirms that the kinetic energy of the alpha particle is 5.38 MeV.

The Balmer series spectral lines in a hydrogen atom result from electron transitions ending at the second energy level, corresponding to $n_1 = 2$. According to the Rydberg formula $\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$ the first spectral line arises from a transition from $n_2 = 3$ to $n_1 = 2$, which gives the inverse wavelength as $\frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$ Similarly, the third spectral line \in this series originates from a transition from the fifth energy level, $n_2 = 5$, to $n_1 = 2$, resulting \in $\frac{1}{\lambda_3} = R \left( \frac{1}{4} - \frac{1}{25} \right) = R \left( \frac{21}{100} \right)$

To determine the ratio of the wavelengths, we divide the expression for the inverse of the third line by the inverse of the first line, which simplifies to $\frac{\lambda_1}{\lambda_3} = \frac{21}{100} \times \frac{36}{5} = 1.512$ Expressing this result \in the form $15.12 × 10^{-1}$ allows us to compare it to the given format $x × 10^{-1}$, leading to the value $x = 15$.

Radioactive decay follows a geometric progression where the activity reduces by half after every half-life interval. Reaching $\frac{1}{16}$ of the initial activity requires the substance to undergo 4 such intervals, as the activity successively becomes $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, and $\frac{1}{16}$ of the starting value. Given that the total time for these 4 half-lives is 80 days, we calculate the half-life using the relationship $4 \times t_{1/2} = 80$ which confirms that the half-life is 20 days.

Radioactive decay and the discharging process of a capacitor both exhibit exponential behavior over time. The activity of a sample follows $A(t) = A_0 e^{-\lambda t}$ while the charge on a discharging capacitor is given by $q(t) = Q_0 e^{-\frac{t}{RC}}$. If the ratio of the charge to the activity is constant, the time-dependent exponential factors in both expressions must cancel out, which requires the decay constant $\lambda$ to be equal to $\frac{1}{RC}$.

Recognizing that the mean life $T_m$ is defined as $\frac{1}{\lambda}$, we can establish the relationship $T_m = RC$, which allows us to determine the resistance using the formula $R = \frac{T_m}{C}$. Inserting the given values of 30 ms for the mean life and 200 µF for the capacitance, the calculation becomes $R = \frac{30 \times 10^{-3}}{200 \times 10^{-6}}$. Completing this calculation yields a resistance of 150 Ω.

The Rydberg formula determines the wavelength of spectral lines by relating the energy transition of an electron to the Rydberg constant and energy levels, defined by $\frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$, where $n_2$ is the final state and $n_1$ is the initial state. For the third member of the Lyman series, the electron transitions to the ground state $n_2 = 1$ from the fourth energy level $n_1 = 4$, which gives $\frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = \frac{15R}{16}$. The first member of the Paschen series involves a transition to the $n_2 = 3$ state from $n_1 = 4$, resulting in $\frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = \frac{7R}{144}$. By dividing the expressions for the wavelengths, $\lambda_1 = \frac{16}{15R}$ and $\lambda_2 = \frac{144}{7R}$, the Rydberg constant cancels out to provide the ratio $\frac{\lambda_1}{\lambda_2} = \frac{16}{15} \times \frac{7}{144} = \frac{7}{135}$.

Atomic spectra serve as unique fingerprints for elements because electrons occupy specific quantized stationary orbits, consistent with Bohr's condition $mvr = \frac{nh}{2\pi}$. Radioactivity is a direct manifestation of nuclear instability, yet the statement claiming nuclear density depends on size is false, as nuclear density is actually a constant property. Finally, the claim regarding particle decay is inaccurate because free protons are stable, whereas free neutrons are the particles that undergo decay.

The energy levels of a hydrogen-like atom or ion are governed by the relationship $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$, where $Z$ is the atomic number and $n$ is the principal quantum number. Since the ground state of a hydrogen atom corresponds to $Z=1$ and $n=1$, its energy is exactly $-13.6 \text{ eV}$. For a single ionized carbon atom, the atomic number is $Z=6$. Setting its energy level equal to the ground state of hydrogen requires solving $-13.6 = -13.6 \frac{6^2}{n^2}$. Dividing both sides by $-13.6$ simplifies the equation to $1 = \frac{36}{n^2}$, implying that $n^2 = 36$, which results in $n=6$.

The energy required to ionize an atom is directly proportional to the mass of the particle orbiting its nucleus. Within the framework of the Bohr model, the ionization potential is defined by the relationship $IP \propto m$, where the mass of the lepton replaces the standard electron mass. Because a muon carries the same electrical charge as an electron, all other physical constants in the energy expression remain fixed. Multiplying the standard hydrogen ionization value of 13.6 eV by the mass ratio of the muon to the electron gives the new potential, $207 \times 13.6$ eV, yielding 2815.2 eV.

The principle of energy conservation requires that the total internal energy lost by the nucleus equals the sum of the emitted photon's energy and the kinetic energy acquired by the recoiling nucleus. Conservation of momentum dictates that the nucleus must recoil with a momentum $p$ equal in magnitude to the photon's momentum, where $p = \frac{hv}{c}$, leading to a recoil kinetic energy of $K = \frac{p^2}{2M} = \frac{(hv)^2}{2Mc^2}$. Summing the photon energy $hv$ with this recoil kinetic energy provides the total decrease in internal energy. Factoring out $hv$ from this sum results in the final expression $hv[1+\frac{hv}{2(Mc)^2}]$.