Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

The Rydberg formula provides the foundation for calculating the wavelengths of spectral lines in hydrogen, defined by the equation $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $n_1$ is the lower energy level and $n_2$ is the upper energy level. Applying this to the first transition line of the Lyman series, where $n_1=1$ and $n_2=2$, yields $\frac{1}{\lambda} = R_H \left( 1 - \frac{1}{4} \right) = \frac{3R_H}{4}$, which simplifies to $\lambda = \frac{4}{3R_H}$.

For the third transition line of the Paschen series, the electron moves from $n_2=6$ to $n_1=3$, resulting in $\frac{1}{\lambda_3} = R_H \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{36} \right) = \frac{3R_H}{36} = \frac{R_H}{12}$, or $\lambda_3 = \frac{12}{R_H}$. Similarly, the second line of the Balmer series involves a transition from $n_2=4$ to $n_1=2$, so $\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R_H}{16}$, which means $\lambda_2 = \frac{16}{3R_H}$.

The problem defines the difference between these two wavelengths as $a\lambda$, expressed as $\frac{12}{R_H} - \frac{16}{3R_H} = a \left( \frac{4}{3R_H} \right)$. Finding a common denominator for the left side gives $\frac{36}{3R_H} - \frac{16}{3R_H} = \frac{20}{3R_H}$. Equating this to the right side, $\frac{20}{3R_H} = a \left( \frac{4}{3R_H} \right)$, leads to the final calculation where $20 = 4a$, confirming that $a = 5$.

Nuclear decay follows strict conservation laws for both mass and atomic numbers throughout the transmutation process. The change in mass number from ${}_{92}^{238}\mathrm{U}$ to ${}_{82}^{206}\mathrm{Pb}$ is $238 - 206 = 32$. Since alpha particles are the only ones carrying significant mass, each reducing the mass number by $4$, dividing $32$ by $4$ confirms that $8$ alpha particles are released. To balance the atomic number, the initial $92$ must equal the sum of the daughter nucleus atomic number $82$, the contribution from the alpha particles $2 \times 8 = 16$, and the effect of beta emission represented by $y$. Setting up the equation $92 = 82 + 16 - y$ accounts for the total charge, which simplifies to $92 = 98 - y$, indicating that $y = 6$ beta particles are emitted.

Bohr's theory establishes that an electron's angular momentum in a hydrogen atom is quantized by the expression $L = \frac{nh}{2\pi}$, where $n$ is the principal quantum number. When a hydrogen atom in its ground state absorbs energy, the electron transitions to a higher energy level, resulting in a change in its angular momentum defined by the difference between the final and initial states.

The energy of the electron in the $n$-th orbit is given by $E_n = \frac{-13.6}{n^2}$ eV. Given the initial ground state energy is -13.6 eV, absorbing 10.2 eV leads to a final energy of -3.4 eV, which corresponds to the state $n=2$. The change in angular momentum is calculated as $\Delta L = L_2 - L_1$, which simplifies to $\Delta L = (2 - 1) \frac{h}{2\pi} = \frac{h}{2\pi}$. Substituting the provided Planck's constant, the calculation becomes $\Delta L = \frac{6.6 \times 10^{-34}}{2 \times 3.14}$, which yields $1.05 \times 10^{-34}$ J s.

The behavior of an electron in a hydrogen atom is governed by energy conservation principles during transitions between orbits. When an electron moves from a lower energy orbit ($E_1$) to a higher energy orbit ($E_2$), it must gain energy by absorbing a photon rather than emitting radiation, which makes the relationship $hf = E_1 - E_2$ physically incorrect. Conversely, when an electron transitions from a higher energy orbit ($E_2$) to a lower energy orbit ($E_1$), the atom releases the excess energy as a photon, satisfying Bohr's frequency condition. The energy difference between these states manifests as the frequency of the emitted photon according to:

$f = \frac{E_2 - E_1}{h}$

Radioactive decay is characterized by the exponential law $A = A_0 e^{-\lambda t}$, where the decay constant $\lambda$ represents the probability of decay per unit time. The average life of the sample, $\tau$, is related to this constant by the inverse relationship $\tau = \frac{1}{\lambda}$, which allows us to determine the sample's lifespan by analyzing its activity at different moments.

To calculate the average life, we express the activity at times $t_1$ and $t_2$ as $A_1 = A_0 e^{-\lambda t_1}$ and $A_2 = A_0 e^{-\lambda t_2}$. By taking the ratio of these two activities, we get $\frac{A}{A/5} = \frac{A_0 e^{-\lambda t_1}}{A_0 e^{-\lambda t_2}}$, which simplifies to $5 = e^{\lambda(t_2 - t_1)}$. Applying the natural logarithm to both sides yields $\ln 5 = \lambda(t_2 - t_1)$, allowing us to determine that $\lambda = \frac{\ln 5}{t_2 - t_1}$. Finally, taking the reciprocal of the decay constant gives the average life as $\tau = \frac{t_2 - t_1}{\ln 5}$.

The frequency of radiation emitted during an electronic transition is directly proportional to the difference in the inverse squares of the principal quantum numbers, described by $f \propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$. Because energy levels in a Bohr atom are spaced closer together as $n$ increases, transitions between the lowest orbits result in the largest energy gaps and consequently the highest frequencies. Evaluating the term $\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ for each choice yields approximately $0.05$ for the transition from $n=4$ to $n=3$, $0.75$ for the $n=2$ to $n=1$ transition, $0.0225$ for the $n=5$ to $n=4$ transition, and $0.14$ for the $n=3$ to $n=2$ transition. Comparing these results, the transition from $n=2$ to $n=1$ produces the greatest value, identifying it as the one with the maximum frequency.

The emission frequency of a hydrogen-like ion is governed by the Rydberg formula, where the frequency is proportional to the product of the Rydberg constant, the square of the atomic number, and the difference of the inverse squares of the principal quantum numbers involved in the transition. Because the constant factors remain identical for both transitions, the ratio of the emitted frequencies depends solely on the energy levels.

For the first transition from $n=3$ to $n=1$, the frequency $f_1$ is proportional to $(1 - 1/3^2)$, which simplifies to $8/9$. The second transition from $n=2$ to $n=1$ yields a frequency $f_2$ proportional to $(1 - 1/2^2)$, which simplifies to $3/4$. Determining the ratio between these frequencies gives $\frac{f_1}{f_2} = \frac{8/9}{3/4} = 32/27$. By substituting the given frequency 2.92 x 10^15 Hz for $f_1$ into the relationship $f_2 = f_1 \times (27/32)$, the resulting frequency is 2.46 x 10^15 Hz.

A free proton cannot spontaneously decay into a neutron because the mass of a neutron is greater than the mass of a proton ($m_n > m_p$), which violates the energy requirements for a particle to decay into a more massive state. This process only becomes viable inside an atomic nucleus, where the energy provided by nuclear binding forces effectively offsets the inherent mass deficit. Consequently, while the transformation is energetically forbidden for a solitary particle, the unique environment within the nucleus permits the conversion.

Spectral series in hydrogen are characterized by the final energy level $n_f$ where the electron lands, while the order of the spectral line depends on the initial state $n_i$, starting from $n_f+1$. Transition $A$ involves a jump from the infinite state down to $n=1$, identifying it as the series limit of the Lyman series. Transition $B$ is a drop from $n=5$ to $n=2$, which constitutes the third member of the Balmer series since the preceding lines originate from $n=3$ and $n=4$. Lastly, transition $C$ originates at $n=5$ and terminates at $n=3$, making it the second member of the Paschen series, as the first line would have started from $n=4$.

To determine how the orbital radius scales with the quantum number, we equate the centripetal force required for circular motion to the negative gradient of the central potential. Given $U(r)=U_{0} r^{4}$, the force magnitude is $F = 4 U_{0} r^{3}$, and equating this to $\frac{m v^{2}}{r}$ implies that $v^{2} \propto r^{4}$, or $v \propto r^{2}$. Substituting this relationship into Bohr's quantization condition $m v r = \frac{n h}{2 \pi}$ leads to $r^{2} \cdot r \propto n$, which reduces to $r^{3} \propto n$. Consequently, $r$ varies with $n^{1/3}$, establishing $\alpha = 3$.

Radioactive decay follows a predictable exponential path where the time required for a substance to reduce its quantity by half remains constant, a duration known as the half-life. When a sample undergoes 33% decay, 67% of the original material remains, and when it undergoes 67% decay, 33% of the original material remains. Calculating the interval between these two states involves determining the time elapsed as the remaining substance decreases from 67% to 33% of its initial value.

The ratio of these two quantities, 0.67 divided by 0.33, is approximately 2. Because the decay constant $\lambda$ relates the time elapsed to the natural logarithm of the ratio of remaining atoms, the time interval $\Delta t$ can be expressed as $\Delta t = \frac{\ln(0.67/0.33)}{\lambda}$. Since this ratio effectively equals 2, the expression simplifies to $\Delta t = \frac{\ln(2)}{\lambda}$, which is the standard definition of the half-life. Therefore, the time interval between these two specific decay states is exactly equal to the half-life of 20 minutes.

Radioactive decay follows the relationship $N = N_0 \left(\frac{1}{2}\right)^n$, where $n$ is the number of half-lives that have elapsed. We are told the half-life of substance $X$ is half the half-life of substance $Y$, expressed as $T_x = \frac{1}{2} T_y$. After a time equal to three half-lives of substance $Y$, substance $Y$ has experienced three half-lives, while substance $X$ has experienced $n = \frac{3 T_y}{T_x} = \frac{3(2 T_x)}{T_x} = 6$ half-lives.

At this instant, the number of undecayed nuclei for both substances is identical. Setting the decay expressions equal gives:
$N_1 \left(\frac{1}{2}\right)^6 = N_2 \left(\frac{1}{2}\right)^3$
Rearranging this expression to isolate the ratio of the initial quantities, we find $\frac{N_1}{N_2} = \frac{(1/2)^3}{(1/2)^6} = 2^3$. This simplifies to a value of 8, meaning the required ratio is 8 to 1.

The wavelength of a photon emitted during an electronic transition within a hydrogen atom is governed by the Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where the Rydberg constant is defined as $R = 1.097 \times 10^7 \mathrm{m}^{-1}$. Substituting the transition from the initial state $n_i = 2$ to the final state $n_f = 1$ into this expression yields the equation $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$. Calculating this value simplifies to $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{3}{4} \right) \mathrm{m}^{-1}$, which leads to a wavelength of $\lambda = 1.215 \times 10^{-7} \mathrm{m}$. Converting this result to nanometers gives 121.5 nm, which aligns with the value of 121.8 nm.

Radioactive decay follows an exponential pattern where the substance reduces to half its quantity during every specific period known as the half-life. If the activity falls to $\left(\frac{1}{8}\right)$ of the initial amount, this signifies that the sample has undergone three successive reductions by half, as $\left(\frac{1}{2}\right)^3$ equals $\frac{1}{8}$. Because it takes three full half-life periods to reach this state, dividing the total time of 30 years by 3 reveals that the half-life of the element is 10 years.

The decay rate of a radioactive sample is determined by its activity, which is the product of its decay constant and the number of atoms present, expressed as $A = \lambda N$. By relating the half-life of element x, defined as $\tau_{1/2} = \frac{\ln(2)}{\lambda_x}$, to the mean life of element y, defined as $\tau = \frac{1}{\lambda_y}$, we find that $\frac{\ln(2)}{\lambda_x} = \frac{1}{\lambda_y}$, which simplifies to $\lambda_x = 0.693 \lambda_y$. Since $\lambda_y$ is greater than $\lambda_x$ and both elements start with the same number of atoms, the activity of element y will be higher, causing it to decay faster than element x.

Radioactive decay is governed by the principle that the number of nuclei in a sample decreases exponentially, reducing by half over every interval defined as the half-life. The remaining amount after a duration $t$ is determined by the expression $N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$, where $N_0$ represents the starting quantity and $T_{1/2}$ is the half-life.

With a starting count of $10^{10}$ nuclei and a half-life of $60 \text{ s}$, an elapsed time of $30 \text{ s}$ corresponds to exactly half of that period. Substituting these values into the expression yields $N = 10^{10} \times \left(\frac{1}{2}\right)^{\frac{30}{60}}$, which simplifies to $N = \frac{10^{10}}{\sqrt{2}}$. Using the provided approximation of $1.414$ for $\sqrt{2}$, the calculation becomes $N \approx \frac{10^{10}}{1.414}$, resulting in $0.707 \times 10^{10}$, or $7 \times 10^{9}$ nuclei remaining.

An X-ray photon emitted as a $K_{\alpha}$ line represents the energy released when an electron drops from the $L$ shell to a vacancy in the $K$ shell, which is described by the equation $E_{K_{\alpha}} = E_K - E_L$. The energy of this photon is determined using the wavelength $\lambda = 0.071 \text{ nm}$ through the relation $E_{K_{\alpha}} = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15} \text{ eVs})(3 \times 10^8 \text{ m/s})}{0.071 \times 10^{-9} \text{ m}}$ which results \in approximately $17.5 \text{ keV}$. Because the $L$-shell vacancy energy $E_L$ is the difference between the $K$-shell energy $E_K$ and the photon energy, it follows that $E_L = 27.5 \text{ keV} - 17.5 \text{ keV}$, giving a final value of $10 \text{ keV}$.

When a radioactive nucleus disintegrates through two distinct and independent pathways, its overall probability of decay per unit time is the combined sum of the individual decay probabilities. This net decay constant is expressed as $\lambda = \lambda_1 + \lambda_2$. Given the fundamental relationship between a half-life and its decay constant, $\lambda = \frac{\ln(2)}{T_{1/2}}$, each individual process can be described by substituting its corresponding half-life into this identity.

Replacing $\lambda$, $\lambda_1$, and $\lambda_2$ with their equivalent expressions in terms of half-lives results in $\frac{\ln(2)}{T_{1/2}} = \frac{\ln(2)}{T_{1/2}^{(1)}} + \frac{\ln(2)}{T_{1/2}^{(2)}}$. Factoring out and canceling the $\ln(2)$ term simplifies the relationship to $\frac{1}{T_{1/2}} = \frac{1}{T_{1/2}^{(1)}} + \frac{1}{T_{1/2}^{(2)}}$. Performing algebraic simplification on this reciprocal sum yields the final expression for the effective half-life:
$T_{1/2} = \frac{T_{1/2}^{(1)} T_{1/2}^{(2)}}{T_{1/2}^{(1)} + T_{1/2}^{(2)}}$

When a radioactive sample decays via two independent channels simultaneously, the total decay constant is the sum of the decay constants of each individual process. The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$, so the combined rate becomes $\lambda_{net} = \frac{\ln 2}{700} + \frac{\ln 2}{1400}$, which simplifies to $\frac{3 \ln 2}{1400}$ per year.

To determine the time $t$ required for one-third of the material to remain, we use the equation $\frac{N_0}{3} = N_0 e^{-\lambda_{net} t}$. This simplifies to the natural log relationship $\ln 3 = \lambda_{net} t$. By substituting the given value $\ln 3 = 1.1$ and the approximation $\ln 2 \approx 0.693$ into the expression $1.1 = \left( \frac{3 \times 0.693}{1400} \right) t$, we find that $t$ is approximately 740 years.

Radioactive decay follows the exponential decay law where the number of remaining nuclei at any time $t$ is expressed as $N(t) = N_0 e^{-\lambda t}$, with $N_0$ representing the initial number of nuclei. To determine the time gap, we first find the times $t_1$ and $t_2$ when the remaining nuclei are $\frac{3 N_0}{4}$ and $\frac{N_0}{2}$ respectively, leading to the relationships $\ln \left( \frac{3}{4} \right) = -\lambda t_1$ and $\ln \left( \frac{1}{2} \right) = -\lambda t_2$. Subtracting these expressions allows us to find the elapsed time $\Delta t = t_2 - t_1$ by calculating $\ln \left( \frac{3}{4} \right) - \ln \left( \frac{1}{2} \right) = \lambda (t_2 - t_1)$, which simplifies to the final result:

$\Delta t = \frac{\ln(3/2)}{\lambda}$