Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

Nuclear density remains constant across all atomic nuclei and is calculated by dividing the total nuclear mass by the nuclear volume, where the radius depends on the mass number through the relation $R = R_0 A^{1/3}$. Expressing the volume explicitly gives $\frac{4}{3}\pi R_0^3 A$, allowing the density equation to be rearranged to solve for the mass number as $A = \frac{M}{\rho} \times \frac{3}{4\pi R_0^3}$. Substituting the given mass, nuclear density, and radius parameter values into this rearranged expression yields an intermediate numerical result. Evaluating this computation leads directly to a mass number of approximately twelve.

The total energy of an electron in a hydrogen atom orbit matches the negative of its kinetic energy, establishing a direct link between energy and the orbital radius through the electrostatic relationship $|E| = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{2r}$. First, convert the given energy magnitude into joules by multiplying the electron-volt value with the standard conversion factor of $1.6 \times 10^{-19} \text{ J}$. Next, rearrange the energy equation to solve for the radius, yielding $r = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{2|E|}$. Substituting the numerical values for Coulomb's constant, the electronic charge, and the converted energy produces a radial distance of approximately $2.1 \times 10^{-10} \text{ meters}$.

Group one elements are characterized by a single valence electron occupying the outermost principal shell of their atomic structure. The Aufbau principle dictates that this outermost electron resides in the lowest angular momentum subshell available for that principal quantum number. Because of this electronic architecture, all alkali metals feature an s subshell as their outermost occupied orbital.

The energy of a photon is directly proportional to the speed of light and Planck's constant, and inversely proportional to its wavelength. Using the relation $E = \frac{hc}{\lambda}$, the energy can be computed directly from the given wavelength and fundamental constants. Substituting the values into the formula gives an energy of approximately $0.95 \times 10^{-24}$ joules, which places it on the order of magnitude of $10^{-24}$ joules.

The nuclear radius scales with the mass number according to the relation $r = r_0 A^{1/3}$, which implies that the nuclear volume is directly proportional to the mass number $A$. Since the total mass of a nucleus is also directly proportional to its mass number, the mass density is calculated by dividing mass by volume. Because both mass and volume scale linearly with $A$, the mass number cancels out, revealing that nuclear density remains entirely independent of the mass number.

The momentum conservation law dictates that the product of mass and velocity for both resulting fragments must be equal, meaning $m_1 v_1 = m_2 v_2$. Since nuclear mass is directly proportional to volume and therefore to the cube of the nuclear radius, the mass ratio can be substituted by the cube of the radius ratio. Rearranging this relationship expresses the radius ratio as the inverse cube root of the given velocity ratio of two, yielding a final nuclear size ratio of one to the cube root of two.

The strong interaction holding nucleons together inside an atomic nucleus operates through exchange mechanisms mediated by mesons. Unlike electrostatic or gravitational fields that extend infinitely while weakening gradually, this binding attraction drops off precipitously to zero outside a tiny fraction of a femtometre. Consequently, nuclear forces remain entirely imperceptible at atomic or macroscopic scales, functioning strictly as short range phenomena.

Charged particles moving through Earth's magnetic field experience a Lorentz force that deflects them depending on the angle between their trajectory and the field lines. Near the magnetic poles, the field lines are nearly vertical, allowing lower-energy particles to penetrate the atmosphere easily. In contrast, near the equator, the field lines are horizontal and perpendicular to incoming particles, creating a stronger barrier that requires cosmic rays to possess significantly greater kinetic energy to reach the surface.

During beta decay, a neutron converts into a proton inside the nucleus, keeping the total mass number constant while increasing the atomic number by one. Because the cadmium nucleus undergoes two successive beta decays, its atomic number increases by a total of two while its mass number remains unchanged at 115. This results in a nuclide with an atomic number of fifty, which corresponds to tin.

The penetrating power of radiation is inversely related to its mass and charge while being directly related to its energy. Gamma rays consist of high-frequency electromagnetic photons carrying no electrical charge and zero rest mass, allowing them to pass through thick barriers of matter with minimal interaction. In contrast, charged alpha and beta particles experience continuous electrostatic interactions with atomic electrons, causing them to lose energy rapidly and stop much sooner.

The rate of nuclear fissions per second is determined by dividing the total operating power by the energy released per individual fission event, expressed as $\text{Rate} = \frac{P}{E}$. Converting the energy release into standard units gives $200 × 1.6 × 10^{-13}$ joules per fission. Dividing the operating power of $5$ watts by this energy value yields a total fission rate of $1.56 × 10^{11}$ per second.

Nuclear fission processes must strictly conserve both total mass numbers and total atomic numbers across the reaction. Summing the initial reactants gives a total mass number of $235 + 1 = 236$ and an atomic number of $92 + 0 = 92$. Subtracting the strontium components and the three neutrons from these totals yields a remaining mass number of $143$ and an atomic number of $54$. Matching these conserved values identifies the missing fission fragment and neutron count.

Radioactive decay reduces the remaining quantity of a substance by half for every elapsed half-life period. The total number of half-lives that pass is determined by dividing the total time elapsed by the duration of a single half-life, yielding $n = \frac{6400}{800} = 8$ half-lives. Substituting this value into the decay fraction formula $\frac{N}{N_0} = \left(\frac{1}{2}\right)^{n}$ gives $\left(\frac{1}{2}\right)^{8}$, which simplifies to $\frac{1}{256}$.

According to Bohr's model of the atom, the radius of a stable electron orbit is derived by equating the electrostatic attraction to the centripetal force and applying the angular momentum quantization condition. Simplifying these fundamental relationships shows that the radius is directly proportional to the square of the principal quantum number, expressed as $r \propto n^2$. Consequently, as the energy level increases, the orbit expands quadratically with the quantum number.

Nuclear forces exhibit the fundamental properties of charge independence and charge symmetry, meaning they are mediated by strong interactions regardless of the electrical charges of the interacting nucleons. Because the strong nuclear force operates with identical strength between two protons, two neutrons, or a proton and a neutron, these interactions are equivalent in magnitude. Consequently, the nuclear forces denoted by $F_{pp}$, $F_{nn}$, and $F_{pn}$ are all equal to each other.

Radioactive decay follows an exponential decrease where the fraction of remaining nuclei depends on the number of half-lives elapsed. For substance A, dividing the total time of 80 minutes by its 20-minute half-life yields 4 half-lives, leaving one-sixteenth of the original nuclei. For substance B, the same time divided by its 40-minute half-life yields 2 half-lives, leaving one-fourth of the original nuclei. Comparing these remaining quantities gives a final ratio of one-sixteenth to one-fourth, which simplifies to $1:4$.

Species that possess the exact same total number of electrons are termed isoelectronic and consequently share identical electronic configurations. Carbon has an atomic number of six, meaning a neutral carbon atom contains six electrons. When nitrogen loses one electron to form a positively charged nitrogen ion, its remaining electron count also drops to six, making it isoelectronic with neutral carbon.

The centripetal force requirement under a constant attractive force coupled with the quantization of angular momentum in the Bohr model dictates how the orbital parameters scale with the quantum number $n$. Since the force is constant, equating it to $\frac{mv^2}{r}$ demonstrates that the orbital radius is directly proportional to the square of the speed through $r \propto v^2$. Combining this dependency with the angular momentum quantization condition $mvr = \frac{nh}{2\pi}$ allows the substitution of radius in terms of velocity. Solving the resulting simultaneous relations yields $v \propto n^{1/3}$ for the speed and $r \propto n^{2/3}$ for the radius.

The electrical neutrality of an atom stems from the precise balance between the positively charged protons in its nucleus and the negatively charged electrons orbiting around it. While atoms of most elements are stable under normal conditions, many radioactive or heavier isotopes undergo spontaneous radioactive decay to achieve a lower energy state. Furthermore, although excited atoms emit characteristic spectra, claiming that every single atom across all elements is uniformly stable is factually incorrect.

The energy of a photon corresponding to any transition is inversely proportional to its wavelength, governed by the relation $\Delta E = \frac{hc}{\lambda}$. By evaluating the energy difference for the second and third transitions, we obtain the equations $\frac{hc}{\lambda_2} = E$ and $\frac{hc}{\lambda_3} = 2E$, respectively. Comparing these two expressions immediately establishes that $\lambda_2 = 2\lambda_3$. For the first transition, the energy gap is $3E/2$, which gives the relation $\frac{hc}{\lambda_1} = \frac{3}{2}E$. Comparing this first expression with the second one yields $3\lambda_1 = 2\lambda_2$, which confirms that $\lambda_2$ is greater than $\lambda_1$.