Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

The wavelength of an emitted photon during an electronic transition relates to the energy change through the relation $\Delta E = \frac{hc}{\lambda}$, which establishes that wavelength is inversely proportional to the energy difference between the orbits. As the upper level $n_2$ increases while transitioning to $n_1 = 2$, the energy gap grows larger, resulting \in progressively shorter wavelengths. Therefore, the smallest energy jump from $n_2 = 3$ to $n_1 = 2$ corresponds to the longest wavelength of 656.3 nm, which pairs A with III, while the increasingly larger energy jumps from $n_2 = 4$, $n_2 = 5$, and $n_2 = 6$ match with the shorter wavelengths of 486.1 nm, 434.1 nm, and 410.2 nm to complete the sequence as B-IV, C-II, and D-I.

Specific heat capacity measures the amount of thermal energy required to raise the temperature of a unit mass of a substance by one degree. Because water possesses an exceptionally high specific heat capacity, it can absorb massive quantities of thermal energy generated within a reactor core without experiencing a drastic increase in temperature. This unique thermodynamic property enables it to transport heat away safely and efficiently, preventing overheating.

Nuclear fission in ${}_{92}^{235}\mathrm{U}$ is triggered more effectively by slow-moving neutrons because they spend sufficient time near the heavy nucleus to be successfully captured, unlike rapid neutrons that tend to scatter away. An alpha particle consists of two protons and two neutrons bound together, which is structurally identical to a doubly ionized helium nucleus. Beta particles are high-speed subatomic particles emitted during radioactive decay, corresponding to either fast-moving electrons or positrons. Conversely, gamma rays are high-frequency electromagnetic waves possessing much shorter wavelengths than X-rays, making the statement regarding longer wavelengths incorrect.

During an alpha emission, the mass number decreases by four and the atomic number decreases by two, transforming ${}_{82}^{290}\mathrm{X}$ into ${}_{80}^{286}\mathrm{Y}$. Emission of a positron decreases the atomic number by one while keeping the mass number unchanged, yielding ${}_{79}^{286}\mathrm{Z}$. Subsequent beta-minus decay converts a neutron into a proton, increasing the atomic number by one to form ${}_{80}^{286}\mathrm{P}$. Finally, electron capture reduces the atomic number by one without altering the mass number, resulting in the final nuclide ${}_{81}^{286}\mathrm{Q}$ with a mass number of 286 and an atomic number of 81.

Hydrogen emission spectral series are classified according to the principal quantum number of the lower energy state where electrons complete their transitions. When electrons cascade down to the fourth energy level from higher orbits, the resulting spectral lines form the Brackett series. This specific series lies within the infrared region of the electromagnetic spectrum.

The activity of a radioactive substance decreases over time according to the decay equation $A = \frac{A_0}{2^n}$, where $n$ is the number of half-lives elapsed. Setting the activity ratio equal to the fraction $\frac{1}{16}$ allows us to determine the required number of half-lives as $n = 4$. Multiplying this number of half-lives by the given half-life duration of 20 minutes yields a total time of 80 minutes.

The energy of an electron in any orbit of a hydrogen-like atom is given by the formula $E_n = \frac{-13.6}{n^2}$. The ground state corresponds to $n = 1$, the first excited state to $n = 2$, and the second excited state to $n = 3$. Substituting $n = 3$ into the energy equation yields an energy of -1.51 eV for the second excited state. Ionization requires supplying enough energy to bring this electron to the continuum where energy is zero, meaning an input of 1.51 eV is necessary.

The radius of any allowed orbit in a hydrogen atom varies with the principal quantum number according to the relation $r_n = r_1 n^2$. For the third orbit, substituting $n = 3$ shows that the radius is nine times larger than the innermost orbit radius. Multiplying the given innermost radius by nine gives $47.7 \times 10^{-11}$ meters, which simplifies to $4.77 \text{ Å}$.

The orbital angular momentum of an electron in a hydrogen atom is quantized according to the relation $mvr = \frac{nh}{2\pi}$. Equating the given angular momentum to this formula gives $\frac{nh}{2\pi} = 1.5 \left(\frac{h}{\pi}\right)$, which directly yields $n = 3$. The energy of an orbit is given by $E_n = -\frac{13.6}{n^2}$ in units of electron volts, resulting in an energy of approximately -1.5 eV for the third orbit.

The reciprocal of wavelength for any spectral series transition is given by the Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$. The shortest wavelength corresponds to the transition from infinity where $n_1 = \infty$. For the Balmer series, the lower level is $n_2 = 2$, yielding a shortest wavelength expression of $\lambda = \frac{4}{R}$, while for the Brackett series, the lower level is $n_2 = 4$, giving $\lambda' = \frac{16}{R}$. Comparing these two expressions shows that the shortest wavelength of the Brackett series is exactly four times that of the Balmer series.

The Lyman series corresponds to electronic transitions ending at the ground state of the hydrogen atom, which involves the largest energy changes among all spectral series. These high-energy transitions emit photons with very short wavelengths ranging from approximately 912 Å to 1216 Å. Since this wavelength range falls well below the visible spectrum, the entire Lyman series resides exclusively in the ultraviolet region.

Radioactive decay follows a statistical law where the rate of disintegration of a sample is directly proportional to the number of undecayed nuclei present at that instant, meaning Statement I is false due to claiming an inverse proportion. Additionally, the half-life of a radionuclide represents the specific duration required for exactly half of the initial active nuclei to decay, rather than a ratio of total lifetimes to initial concentration as claimed in Statement II. Since both definitions fail to accurately describe these fundamental radioactive principles, both statements are incorrect.

The energy of an electron in a hydrogen atom varies inversely with the square of the principal quantum number according to the relation $E_n = \frac{E_0}{n^2}$. The ground state corresponds to $n = 1$, making the first excited state $n = 2$ and the second excited state $n = 3$. Substituting these quantum numbers into the energy formula gives $T_1 = \frac{E_0}{2^2}$ and $T_2 = \frac{E_0}{3^2}$. Taking the ratio of $T_1$ to $T_2$ results in the fraction $\frac{T_1}{T_2} = \frac{3^2}{2^2}$, which simplifies to $\frac{9}{4}$.

The radius of a stationary orbit in the Bohr model is directly proportional to the square of the principal quantum number, following the proportionality $R \propto n^2$. For the second stationary orbit the quantum number is $n = 2$, and for the fourth stationary orbit it is $n = 4$. Taking the ratio of these radii gives $\frac{R_1}{R_2} = \frac{2^2}{4^2}$, which simplifies to $\frac{4}{16}$ or $0.25$.

The radius of a nucleus scales directly with the cube root of its mass number according to the relation $R = R_0 A^{1/3}$. Applying this scaling law to the two daughter nuclei with mass numbers 125 and 64 allows us to set up their radius ratio as $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$. Substituting the given values yields $\left(\frac{125}{64}\right)^{1/3}$, which evaluates cleanly to $\frac{5}{4}$.

The number of remaining radioactive atoms at any time follows the exponential decay law $N = N_0 e^{-\lambda t}$. Setting up the ratio of the atoms for both elements gives $\frac{N_{1}}{N_{2}} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}}$, which simplifies to $e^{-9\lambda t}$. Equating this ratio to $\frac{1}{e}$, the exponent comparison yields $9\lambda t = 1$, resulting in the required time $\frac{1}{9\lambda}$.

Nuclear reactions must satisfy the conservation of mass number and the conservation of atomic number on both sides of the equation. Equating the mass numbers gives $22 = A + 0$, which yields a mass number $A = 22$ for the unknown nucleus. Similarly, equating the atomic numbers gives $11 = Z + 1$, resulting in an atomic number $Z = 10$. The element with atomic number 10 is neon, meaning the resulting nucleus is represented as ${}_{10}^{22}\mathrm{Ne}$.

Radioactive transformations alter the atomic number of a nucleus in a specific manner depending on the emitted particle. Specifically, positron emission reduces atomic number by $1$, alpha decay reduces it by $2$, and electron emission increases it by $1$. Tracking these modifications from $Z$ through the intermediate nuclei directly identifies the decay sequence as $\beta^+$, $\alpha$, and $\beta^-$.

The total binding energy of a nucleus is found by multiplying its mass number by the binding energy per nucleon, meaning the energy gain is the difference between the total binding energy of the products and that of the reactant. Calculating the product binding energy gives $240 \times 8.5$ MeV, and calculating the reactant binding energy gives $240 \times 7.6$ MeV. Subtracting the initial energy from the final energy yields a net binding energy gain of $216$ MeV.

The activity of a radioactive sample diminishes over time following the relation $A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$. Substituting the given time of $150$ hours and half-life of $100$ hours into the exponent gives the ratio $\left(\frac{1}{2}\right)^{\frac{150}{100}}$. Simplifying the fractional exponent to $\frac{3}{2}$ yields $\left(\frac{1}{2}\right)^{\frac{3}{2}}$, which evaluates to $\frac{1}{2\sqrt{2}}$.