Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

The wavelength of light emitted during an electronic transition is determined using the Rydberg relation $\frac{1}{\lambda} = R \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right)$. Inserting the final orbit value of 2 and the initial orbit value of 4 into the equation gives $R \left(\frac{1}{4} - \frac{1}{16}\right)$. Simplifying the bracketed fractions results in $R \left(\frac{3}{16}\right)$ for the inverse wavelength. Inverting this expression provides the final wavelength value of $\frac{16}{3 R}$.

An absorption spectrum exclusively involves the excitation of electrons from the ground level to various higher energy levels. Consequently, only the transitions starting directly from the lowest ground state will be observed, which means lines 1, 2, and 3 are the ones that will occur in the absorption spectrum.

Line emission spectra are produced by low-pressure gases when electric current excites their atoms, causing them to emit light at specific, discrete wavelengths as electrons drop to lower energy states. Incandescent sources produce continuous spectra, whereas gas-discharge tubes contain rarified atomic gas that radiates characteristic sharp spectral lines unique to their elemental composition.

Radioactive decay causes activity to decrease exponentially over time, which means moving backward requires scaling the activity up by a factor of two for each elapsed half-life. Since the time gap of two months spans exactly two half-lives given the one-month half-life period, the earlier activity must be four times larger than the current reading. Multiplying the given two microcuries by this factor yields an initial activity of eight microcuries.

The atomic number of an element defines its proton count, while the mass number represents the combined total of protons and neutrons within the nucleus. For sodium with an atomic number of 11 and a mass number of 23, the nucleus contains 11 protons and 12 neutrons derived from their difference. Because electrons exist exclusively in the outer atomic orbitals and are completely absent from the nuclear region, the number of electrons located inside the nucleus is zero.

The rate at which an unstable atomic nucleus undergoes radioactive disintegration is quantified using specific activity units. The curie was originally established to measure the activity level corresponding to one gram of radium, representing a specific number of nuclear disintegrations per second. This metric serves strictly to express the intensity of decay events occurring within a radioactive sample rather than measuring energy levels or half-life durations.

Nuclear fission reactions produce high-speed neutrons that are too fast to efficiently cause subsequent fissions in uranium nuclei. To sustain the chain reaction, these fast neutrons must be slowed down to thermal energies by colliding with light nuclei in a substance called a moderator. Materials like heavy water and graphite excel at this because their constituent nuclei absorb very little neutron flux while effectively reducing neutron velocity through elastic collisions.

Nuclear stability is directly indicated by the binding energy per nucleon curve, which plots how tightly nucleons are bound together across different elements. This curve climbs sharply from light nuclides and reaches its global maximum in the vicinity of medium-mass elements. Iron nuclei occupy the absolute peak of this curve because their constituent nucleons achieve the highest possible binding energy per particle. Consequently, iron possesses the strongest binding energy per nucleon among all elements.

Nuclear radius scales proportionally with the cube root of the mass number according to the empirical relation $R = R_0 A^{1/3}$. Applying this scaling law to aluminum-27 gives a proportional radius factor of $(27)^{1/3}$, which simplifies to $3$. Doing the same for tellurium-125 yields a proportional radius factor of $(125)^{1/3}$, which simplifies to $5$. Taking the ratio of these two radius values results in $\frac{3}{5}$, which expands equivalently to $\frac{6}{10}$.

Radioactive decay follows a geometric progression where the remaining fraction of a substance after a given duration depends on the number of half-lives that have elapsed. The total elapsed time divided by the half-life period determines this exponent, which evaluates to 6400 divided by 1600, resulting in 4 half-lives. Raising the base fraction of one-half to the power of 4 yields the final remaining fraction of $\left(\frac{1}{2}\right)^{4} = \frac{1}{16}$.

Radioactive decay processes alter atomic and mass numbers systematically, where alpha emission decreases both numbers while beta emission increases the atomic number without affecting the mass number. In the initial alpha decay step, the parent nucleus loses two protons and two neutrons, reducing its atomic number by two and its mass number by four. The subsequent dual beta emission increases the atomic number by two while leaving the mass number completely unchanged, meaning the final nucleus shares the original atomic number but has a smaller mass number, which classifies the initial and final elements as isotopes.

The total binding energy of a nucleus is calculated by multiplying its total number of nucleons by the binding energy per nucleon. On the reactant side, two deuterons combine to provide a total of four nucleons each bound with an energy of $x_1$, giving a total reactant binding energy of $4x_1$. On the product side, the resulting alpha particle consists of four nucleons bound with an energy of $x_2$, resulting in a total product binding energy of $4x_2$. Subtracting the initial binding energy from the final binding energy determines the net energy released during the reaction as $4(x_2 - x_1)$.

Nuclear fusion is the process in which light atomic nuclei combine under extreme temperature and pressure to form heavier nuclei, releasing a massive amount of energy in the process. Inside the core of the sun, immense gravitational forces drive the continuous fusion of hydrogen nuclei into helium. This stellar reaction converts a small fraction of mass directly into radiant energy, powering the sun and emitting light across the solar system.

The energy levels of a hydrogen atom are quantized and given by the formula $E_n = -\frac{13.6}{n^2}$ eV. For the first excited state, the principal quantum number is $n = 2$, which places the electron in the second orbit. Substituting this value into the equation yields $E_2 = -\frac{13.6}{2^2} = -3.4$ eV.

The dependence of deflected alpha particles on the deflection angle in the Geiger-Marsden experiment is governed by the relation $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$ which dictates a dramatic reduction in particle count as the angle increases. Because of this inverse fourth-power relationship, an enormous volume of alpha particles experiences almost no deviation at tiny angles, yielding a very high initial peak on the graph. As the angle expands past ninety degrees and approaches one hundred eighty degrees, the curve exhibits a steep, precipitous drop toward nearly zero, reflecting the extreme rarity of head-on collisions with the atomic nucleus.

The stability of electron orbits in the Bohr model is maintained by proposing that the orbital angular momentum of an electron is quantized. Specifically, this restriction dictates that angular momentum can only take discrete values equal to integral multiples of $\frac{h}{2\pi}$, ensuring that electrons do not continuously radiate energy while occupying stationary states.

The emission of an alpha particle reduces the mass number by four units, whereas beta emissions leave the mass number completely unchanged while increasing the atomic number. The total change in mass number from the reactant to the product is found by subtracting the final mass number from the initial mass number, yielding thirty-two. Dividing this total mass change by four gives the total number of alpha particles emitted as eight, and accounting for the subsequent atomic number adjustments reveals six emitted beta particles.

The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ \mathrm{eV}$. To find the energy change during a transition, we calculate the energy for the fourth and third orbits separately using this expression. Substituting the quantum numbers yields $E_4 = -0.85 \ \mathrm{eV}$ and $E_3 = -1.51 \ \mathrm{eV}$. The energy difference corresponding to the transition between these two orbits is therefore given by $\Delta E = E_4 - E_3$, which evaluates to $-0.85 \ \mathrm{eV} - (-1.51 \ \mathrm{eV}) = 0.66 \ \mathrm{eV}$.

Nuclear transmutation follows the conservation of mass and atomic numbers during particle absorption and decay. Adding a neutron to carbon-12 produces carbon-13 through the equation ${}_{6}\mathrm{C}^{12} + {}_{0}\mathrm{n}^{1} \rightarrow {}_{6}\mathrm{C}^{13}$. The subsequent beta emission converts a neutron into a proton according to ${}_{6}\mathrm{C}^{13} \rightarrow {}_{7}\mathrm{N}^{13} + {}_{-1}\beta^{0}$, which raises the atomic number to seven while keeping the mass number at thirteen.

Electrons fill atomic orbitals starting from the lowest available energy levels up to the maximum capacity of each subshell until all fourteen electrons of silicon are accommodated. This sequential filling populates the first and second principal energy levels completely with ten electrons, while the remaining four electrons populate the third shell across the $3s$ and $3p$ subshells.