Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

The energy released in a nuclear reaction, designated as $Q$, is calculated using the energy balance relation $Q = E_{\text{products}} - E_{\text{reactants}}$. Multiplying the number of nucleons in each nucleus by its respective binding energy per nucleon gives the total binding energies for both the helium products and the lithium reactant. Subtracting the reactant binding energy from the combined product binding energy yields a released energy value of 17.3 MeV.

The energy of an incident photon determines the excitation level of a hydrogen atom by matching the difference between energy states, and the total number of spectral lines emitted during subsequent de-excitation depends on the highest excited state reached. Calculating the photon energy from the given wavelength yields 12.75 electron-volts, which excites a ground-state hydrogen atom up to the fourth principal shell. Applying the combinatorial formula for transitions from this level, given by $\frac{n(n-1)}{2}$, results in a total of 6 spectral lines.

Penetrating power through a medium depends on the mass, charge, and ionizing capability of radiation, where particles with higher mass and charge interact more strongly and thus penetrate less. Alpha particles possess the largest mass and charge, making them highly ionizing and easily stopped, whereas gamma rays are neutral photons with high energy and minimal direct interaction. Consequently, for a given energy level, penetrating power increases progressively from alpha particles through beta particles to gamma rays.

The energy released in a nuclear fusion reaction is determined by multiplying the mass defect by the energy equivalent of an atomic mass unit, utilizing the relation $E = \Delta M \times c^2$. Given a mass defect of 0.02866 u and an energy equivalent of 931 MeV per u, the total energy is calculated as their product. Dividing this total energy by the four nucleons participating in the reaction yields the energy liberated per u, which evaluates to 6.675 MeV.

The longest wavelength in any spectral series of hydrogen is obtained from the Rydberg formula $\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ by considering the transition from the immediately adjacent upper energy level. For the Lyman series, this transition spans from the second level to the first, whereas for the Balmer series, it spans from the third level to the second. Computing the inverse wavelengths for these respective transitions and taking their ratio yields $\frac{5}{27}$.

Nuclear stability is directly reflected by the binding energy per nucleon curve, which rises sharply for light nuclei as the strong nuclear force quickly overcomes surface effects. This curve reaches a maximum plateau around a mass number of fifty-six where nuclei are most tightly bound and stable. Beyond this threshold, the escalating electrostatic repulsion between protons gradually outpaces the short-range attractive nuclear force, causing the binding energy per nucleon to decrease steadily for heavier elements.

The age of a radioactive sample is calculated by determining how many half-lives have elapsed using the decay relation $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$. Given that the ratio of the remaining isotope X to the stable decay product F is $1:7$, the fraction of the remaining isotope relative to the initial amount is $\frac{1}{1 + 7}$, which simplifies to $\frac{1}{8}$. Equating this fraction to $\left(\frac{1}{2}\right)^n$ reveals that $n = 3$ half-lives have passed. Multiplying this number of half-lives by the given half-life of 20 years gives the total age of the rock as 60 years.

According to Bohr model derivations, the time period of an electron in a stationary orbit is proportional to the cube of the principal quantum number, expressed as $T \propto n^3$. Taking the ratio of the initial to final time periods yields an eightfold relationship, which equals the cube of the ratio of their respective quantum numbers. Taking the cube root establishes that the initial quantum number is twice the final one, matching the values 4 and 2.

The wavelength of the emitted photon is obtained from the Rydberg formula $\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$, where substituting the initial and final energy levels yields the fractional wave number. Applying the conservation of linear momentum, the momentum of the emitted photon given by $\frac{h}{\lambda}$ equals the recoil momentum of the atom $mv$. Substituting this momentum relation into the wave number expression yields the recoil velocity as $\frac{24hR}{25m}$.

Nuclear radius scales with the mass number according to the relation $R = R_0 A^{1/3}$, where $R_0$ is a proportionality constant and $A$ is the mass number. Setting up a ratio of the radii for aluminum and copper eliminates the constant and equates it to the cube roots of their respective mass numbers. Substituting mass numbers 27 and 64 evaluates this ratio, and multiplying by the given aluminum radius of 3.6 fermi results in 4.8 fermi for copper.

Photon energy emitted during an electronic transition is inversely proportional to the wavelength of the radiation, meaning smaller energy jumps yield longer wavelengths such as infrared light. Transitions ending at the ground state produce high-energy ultraviolet photons, whereas those terminating at higher principal quantum numbers involve much smaller energy gaps. Among the given choices, the transition from the fourth to the third energy level belongs to the Paschen series, which characterizes the infrared region of the spectrum.

Applying the radioactive decay law $N = N_0 e^{-\lambda t}$, the number of remaining nuclei at times $t_2$ and $t_1$ can be expressed as one-third and two-thirds of the initial amount, respectively. Dividing the two resulting decay equations eliminates the initial population and yields the expression $\lambda (t_2 - t_1) = \ln(2)$. Substituting the decay constant relation $\lambda = \frac{\ln(2)}{T_{1/2}}$ directly isolates the time interval as equal to the half-life of fifty days.

The wavelengths associated with electron transitions in a hydrogen atom are determined using the Rydberg formula $\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$. For the transition from the third excited state to the second excited state, and subsequently from the second to the first excited state, the inverse wavelengths are calculated for each respective pair of initial and final energy levels. Taking the ratio of the resulting wavelengths yields $\frac{27}{5}$.

The remaining quantity of a radioactive substance after a given duration is expressed by the decay equation $N = N_0\left(\frac{1}{2}\right)^{t/T}$, where $N_0$ is the initial amount and $T$ is the half-life. Setting up the respective decay expressions for both materials and equating them accounts for the equal amounts at time $t$. Simplifying the resulting exponential equation isolates time and yields 40 s upon solving the linear relation.

The energy levels in a hydrogen atom are quantized and given by the formula $E_n = -13.6 / n^2$, meaning emitted photons must match the exact energy difference between two discrete orbital states. Evaluating this formula for the first few orbits gives ground state energy as -13.6 eV, the second level as -3.4 eV, the third level as -1.5 eV, and the fourth level as -0.85 eV. The transition between the fourth and third levels yields 0.65 eV, the transition between the third and second levels yields 1.9 eV, and a drop to the ground state can release 13.6 eV. Because 11.1 eV cannot be produced by finding the difference between any possible pair of these energy levels, it is not a valid photon emission energy.

The emission of a photon carries momentum, and by the law of conservation of linear momentum, the recoiling nucleus must acquire an equal and opposite momentum. Equating the photon momentum formula $p = \frac{h\upsilon}{c}$ to the momentum of the nucleus $Mv$ allows us to solve for the recoil velocity as $v = \frac{h\upsilon}{Mc}$. Substituting this velocity into the standard kinetic energy equation $E = \frac{1}{2}Mv^2$ produces the final expression for the recoil energy as $\frac{h^2\upsilon^2}{2Mc^2}$.

Nuclear fusion requires extreme thermal energy because positively charged nuclei naturally experience strong electrostatic repulsion that must be overcome by high kinetic energy to allow the strong nuclear force to bind them together. At ordinary temperatures, this Coulomb barrier prevents light nuclei from getting close enough to fuse. Raising the temperature increases the thermal velocity of the particles sufficiently for them to collide with enough force to bridge this repulsive barrier.

Equating the reciprocal wavelengths of the respective spectral transitions using the Rydberg formula gives $R \left(\frac{1}{1^2} - \frac{1}{2^2}\right)$ for the first Lyman line of hydrogen and $Z^2 R \left(\frac{1}{2^2} - \frac{1}{4^2}\right)$ for the second Balmer line of the ion. Simplifying both sides of this equality results in $\frac{3}{4} = \frac{3 Z^2}{16}$. Solving this algebraic expression immediately yields an atomic number of two for the hydrogen-like ion.

Radioactive decay follows an exponential decay law where the remaining number of nuclei at any given time depends on the initial quantity and the half-life of the substance. Equating the population formulas for both species allows us to determine the exact time elapsed when their quantities become equal, which occurs after 4 minutes. Subtracting the remaining undecayed nuclei of both elements at this instant from their respective initial amounts gives the total number of nuclei that have transformed into the stable product R, yielding $9N_0 / 2$.

Radioactive decay follows specific conservation laws where alpha emission decreases the mass number by 4 and atomic number by 2, while beta-minus emission increases the atomic number by 1 without altering the mass number. When the initial nucleus $^{m}_{n}\mathrm{X}$ emits an alpha particle, the resulting intermediate nucleus has a mass number of $m-4$ and an atomic number of $n-2$. Subsequent emission of two beta-minus particles increases this atomic number by 2 while leaving the mass number unchanged, resulting in a final nucleus with mass number $m-4$ and atomic number $n$.