Atomic Structure MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Identify the initial binding energy:}\) For hydrogen, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] The atoms initially occupy \(n=2\). Therefore, \[ E_2=-\frac{13.6}{4}=-3.40\,\mathrm{eV} \] The ionisation energy from this state is \(3.40\,\mathrm{eV}\). \(\textbf{Analyse the first absorption in experiment P:}\) The first photon supplies \(2.55\,\mathrm{eV}\). The new atomic energy is \[ E_f=E_2+E_{\mathrm{photon}} \] \[ E_f=-3.40+2.55 \] \[ E_f=-0.85\,\mathrm{eV} \] For hydrogen, \[ E_4=-\frac{13.6}{16}=-0.85\,\mathrm{eV} \] Thus, the first photon excites the atom from \(n=2\) to \(n=4\). \(\textbf{Analyse the second absorption in experiment P:}\) The ionisation energy from \(n=4\) is \[ E_{\mathrm{ion},4}=0-E_4=0.85\,\mathrm{eV} \] The second photon carries \(4.00\,\mathrm{eV}\), which exceeds this threshold. The released-electron kinetic energy is \[ K_P=4.00-0.85 \] \[ K_P=3.15\,\mathrm{eV} \] \(\textbf{Analyse the direct ionisation in experiment Q:}\) The second atom begins in \(n=2\) and absorbs only a \(4.00\,\mathrm{eV}\) photon. Its ionisation energy is \(3.40\,\mathrm{eV}\). Therefore, \[ K_Q=4.00-3.40 \] \[ K_Q=0.60\,\mathrm{eV} \] \(\textbf{Verify the energy accounting:}\) The total photon energy absorbed in experiment P is \[ 2.55+4.00=6.55\,\mathrm{eV} \] Relative to the initial \(n=2\) state, ionisation requires \(3.40\,\mathrm{eV}\). Hence, \[ K_P=6.55-3.40=3.15\,\mathrm{eV} \] This agrees with the two-stage calculation. Experiment Q receives only \(4.00\,\mathrm{eV}\), leaving \(0.60\,\mathrm{eV}\) as electron kinetic energy. The initial bound-state excitation in experiment P changes the state from which the electron is subsequently ionised. The problem therefore requires retaining the atom's intermediate electronic state rather than applying the original ionisation energy separately to both photons. \(\textbf{Final answer:}\) \[ \boxed{K_P=3.15\,\mathrm{eV},\qquad K_Q=0.60\,\mathrm{eV}} \]

Explanation: \(\textbf{Establish the orbit-radius change:}\) For hydrogen in Bohr's model, \[ r_n=a_0n^2 \] Therefore, \[ \frac{r_4}{r_2}=\frac{4^2}{2^2} \] \[ \frac{r_4}{r_2}=4 \] The fourth orbit has four times the radius of the second orbit. \(\textbf{Determine the electrostatic-force ratio:}\) The magnitude of the attractive Coulomb force is \[ F_n=\frac{e^2}{4\pi\varepsilon_0r_n^2} \] For a fixed nuclear charge, force varies inversely with the square of orbital radius. Thus, \[ \frac{F_4}{F_2}=\left(\frac{r_2}{r_4}\right)^2 \] \[ \frac{F_4}{F_2}=\left(\frac14\right)^2 \] \[ \frac{F_4}{F_2}=\frac1{16} \] \(\textbf{Determine the electron-speed ratio:}\) For hydrogen, \[ v_n\propto\frac1n \] Therefore, \[ \frac{v_4}{v_2}=\frac24 \] \[ \frac{v_4}{v_2}=\frac12 \] The electron has half its original speed in the fourth orbit. \(\textbf{Calculate the revolution-period ratio:}\) The time period of circular motion is \[ T_n=\frac{2\pi r_n}{v_n} \] Consequently, \[ \frac{T_4}{T_2}=\frac{r_4}{r_2}\frac{v_2}{v_4} \] Substituting, \[ \frac{T_4}{T_2}=4\times2 \] \[ \frac{T_4}{T_2}=8 \] \(\textbf{Physical interpretation:}\) The electron moves along a larger circular orbit and also has a smaller orbital speed. Both effects increase its revolution period. The electrostatic attraction becomes weaker because of the larger electron-nucleus separation. The force ratio and time-period ratio therefore change by different factors. The orbital revolution period should not be confused with the period of radiation absorbed during the electronic transition. In Bohr's model, the absorbed photon frequency is determined by the difference between stationary-state energies, not simply by the reciprocal of either orbital revolution period. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_4}{F_2}=\frac1{16},\qquad\frac{T_4}{T_2}=8} \]

Explanation: \(\textbf{Determine the initial and final quantum numbers:}\) For a hydrogen atom, the radius of the \(n\)th Bohr orbit is \[ r_n=n^2a_0 \] The initial radius is \(25a_0\). Therefore, \[ n_i^2a_0=25a_0 \] \[ n_i=\sqrt{25}=5 \] The final radius is \(9a_0\). Similarly, \[ n_f^2a_0=9a_0 \] \[ n_f=\sqrt9=3 \] Thus, the electron undergoes the downward transition \[ 5\rightarrow3 \] \(\textbf{Calculate the electronic energies:}\) The Bohr energy expression for hydrogen is \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] For the initial state, \[ E_5=-\frac{13.6}{25} \] \[ E_5=-0.544\,\mathrm{eV} \] For the final state, \[ E_3=-\frac{13.6}{9} \] \[ E_3\approx-1.511\,\mathrm{eV} \] \(\textbf{Determine the emitted photon energy:}\) During a downward transition, the atom loses energy and emits a photon. The emitted photon energy is the positive difference between the initial and final electronic energies: \[ E_{\mathrm{photon}}=E_i-E_f \] Substituting, \[ E_{\mathrm{photon}}=-0.544-(-1.511) \] \[ E_{\mathrm{photon}}\approx0.967\,\mathrm{eV} \] The positive photon energy is consistent with emission from a higher-energy state to a lower-energy state. \(\textbf{Identify the spectral series:}\) Hydrogen emission series are classified according to the final principal quantum number. Transitions terminating at \(n=1\) belong to the Lyman series. Transitions terminating at \(n=2\) belong to the Balmer series. Transitions terminating at \(n=3\) belong to the Paschen series. Since the final state is \(n_f=3\), the transition belongs to the Paschen series. The corresponding radiation lies in the infrared region. \(\textbf{Interpret the physical change:}\) The radius decreases from \(25a_0\) to \(9a_0\), indicating that the electron moves closer to the nucleus in the Bohr model. Its energy becomes more negative, corresponding to a more strongly bound state. The loss of atomic energy appears as the energy of the emitted photon. \(\textbf{Final conclusion:}\) The transition is \(5\rightarrow3\), the emitted photon energy is approximately \(0.967\,\mathrm{eV}\), and the spectral line belongs to the Paschen series.

Explanation: \(\textbf{Identify the common quantum number:}\) Both ions are one-electron species occupying \(n=2\). Their nuclear charges differ: \[ Z_{\mathrm{He}}=2,\qquad Z_{\mathrm{Be}}=4 \] The comparison must therefore retain the same \(n\) while changing \(Z\). \(\textbf{Compare electronic kinetic energies:}\) For a hydrogen-like Bohr orbit, \[ K_n=13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] At fixed \(n\), kinetic energy is proportional to \(Z^2\). Therefore, \[ \frac{K_{\mathrm{Be}}}{K_{\mathrm{He}}}=\frac{4^2}{2^2} \] \[ \frac{K_{\mathrm{Be}}}{K_{\mathrm{He}}}=4 \] Hence, \[ K_{\mathrm{Be}}=4K_{\mathrm{He}} \] \(\textbf{Compare orbital radii:}\) The Bohr radius expression is \[ r_n=a_0\frac{n^2}{Z} \] At fixed \(n\), radius is inversely proportional to nuclear charge. Thus, \[ \frac{r_{\mathrm{Be}}}{r_{\mathrm{He}}}=\frac{Z_{\mathrm{He}}}{Z_{\mathrm{Be}}} \] \[ \frac{r_{\mathrm{Be}}}{r_{\mathrm{He}}}=\frac24 \] \[ r_{\mathrm{Be}}=\frac12r_{\mathrm{He}} \] \(\textbf{Compare orbital angular momenta:}\) Bohr's angular-momentum quantisation condition is \[ L_n=\frac{nh}{2\pi} \] At the same principal quantum number, this angular momentum is independent of \(Z\). Therefore, \[ L_{\mathrm{Be}}=L_{\mathrm{He}} \] \(\textbf{Resolve the apparent contrast:}\) The electron in \(\mathrm{Be^{3+}}\) has greater kinetic energy and therefore greater orbital speed. At the same time, its orbit is smaller because the nuclear attraction is stronger. These changes compensate so that the quantised angular momentum remains equal to that of the electron in \(\mathrm{He^+}\). Thus, equal Bohr angular momentum does not imply equal orbital radius, speed or kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{\frac{K_{\mathrm{Be}}}{K_{\mathrm{He}}}=4,\qquad\frac{r_{\mathrm{Be}}}{r_{\mathrm{He}}}=\frac12,\qquad\frac{L_{\mathrm{Be}}}{L_{\mathrm{He}}}=1} \]

Explanation: The Bohr energy formula in the question is derived for a hydrogen-like system containing one electron attracted to a nucleus. Its direct applicability requires a single-electron species, not merely an atom or ion with a positive nuclear charge. Case P is valid because a neutral hydrogen atom contains one proton and one electron. The quantitative Bohr relations apply to the hydrogen atom within the model. Case Q is valid because \(\mathrm{He^+}\) contains two protons in its nucleus but only one bound electron. Its electronic energy levels therefore follow the hydrogen-like \(Z^2/n^2\) dependence. Case R is not valid for the stated direct treatment because \(\mathrm{Li^+}\) contains two electrons. Lithium has atomic number three, and removal of one electron leaves two bound electrons. The electrons interact not only with the nucleus but also with one another. These electron-electron interactions are not represented in the simple one-electron Coulomb potential used to derive the hydrogenic Bohr energy formula. Case S is valid because \(\mathrm{Be^{3+}}\) has atomic number four and has lost three electrons. It therefore retains only one bound electron. The correct distinction is between nuclear charge and electron count. A larger nuclear charge does not invalidate the hydrogen-like expression when exactly one electron remains. Conversely, an ion with a positive charge does not necessarily become hydrogen-like if it still contains multiple electrons. Therefore, the formula is directly applicable to P, Q and S.

Explanation: Statement I is correct because Bohr's model assigns definite energies to permitted electronic states in the hydrogen atom. Radiation is emitted or absorbed when the electron undergoes a transition between these states. The energy difference determines the photon frequency, allowing the principal hydrogen spectral wavelengths to be predicted. Statement II is incorrect because the simple Bohr treatment assumes a single electron moving under the attraction of a nucleus. Neutral multi-electron atoms contain electron-electron interactions that cannot be described by merely inserting a larger nuclear charge into the hydrogenic formula. The model therefore does not provide the complete detailed spectra of arbitrary neutral atoms. Statement III is correct because the elementary Bohr model cannot fully account for experimentally observed fine spectral details. The splitting of spectral lines in an applied magnetic field is associated with the Zeeman effect. Splitting in an applied electric field is associated with the Stark effect. A complete treatment of these observations requires a more advanced quantum description than the elementary circular-orbit model. Statement IV is incorrect because agreement with selected spectral wavelengths does not prove that a classical electron trajectory has been directly observed. Bohr's orbits are part of a historical model that successfully reproduces important hydrogen-like energy relationships. The modern quantum description does not represent an atomic electron as travelling along a precisely defined classical circular path. Thus, the correct conclusion recognises both the important spectroscopic success and the limitations of the Bohr model. Only Statements I and III are correct.

Explanation: Statement I is correct as a description of Bohr's historical atomic model. For a hydrogen-like species, each permitted principal quantum number corresponds to a definite model radius and electron speed. These values follow from the quantised angular momentum condition combined with the electrostatic force equation. Statement II is incorrect because successful prediction of selected spectral wavelengths does not establish consistency with every principle of modern quantum mechanics. A classical circular trajectory assigns the electron a precisely specified path together with a definite momentum along that path. Heisenberg's uncertainty principle does not permit arbitrarily precise simultaneous knowledge of position and the corresponding momentum. The fixed-trajectory description is therefore incompatible with a complete quantum mechanical description of an electron. Statement III is correct because an atomic orbital is associated with the quantum mechanical wave function of the electron. The orbital description gives information about the spatial probability distribution rather than specifying a classical route travelled by the electron. Statement IV is incorrect because replacing a Bohr radius with a more precisely measured radius does not convert a circular orbit into a quantum mechanical orbital. The two descriptions differ conceptually, not merely in experimental precision. Bohr's model remains useful for understanding hydrogen-like energies and historical quantisation, but its geometrical electron-path picture is not the modern description. The appropriate distinction is between a model trajectory and a probability-based quantum state. Only Statements I and III are correct.

Explanation: \(\textbf{Connect potential and total electronic energies:}\) For an electron in a Coulomb-bound Bohr orbit, \[ U=-2K \] The total energy is \[ E=K+U \] Substituting \(U=-2K\), \[ E=-K \] Therefore, \[ U=2E \] \(\textbf{Determine the total energy of the occupied state:}\) The given electrostatic potential energy is \[ U=-1.70\,\mathrm{eV} \] Hence, \[ E=\frac{U}{2} \] \[ E=\frac{-1.70}{2} \] \[ E=-0.85\,\mathrm{eV} \] \(\textbf{Identify the principal quantum number:}\) For hydrogen, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] Equating this to the measured total energy, \[ -\frac{13.6}{n^2}=-0.85 \] Therefore, \[ n^2=\frac{13.6}{0.85} \] \[ n^2=16 \] \[ n=4 \] \(\textbf{Calculate the kinetic energy:}\) Since \(K=-E\), \[ K=0.85\,\mathrm{eV} \] This agrees with the alternative calculation \[ K=-\frac{U}{2} \] \[ K=-\frac{-1.70}{2}=0.85\,\mathrm{eV} \] \(\textbf{Determine the ionisation energy:}\) The electron is bound at total energy \(-0.85\,\mathrm{eV}\). The reference energy of a separated electron at rest is zero. Therefore, \[ E_{\mathrm{ion}}=0-E \] \[ E_{\mathrm{ion}}=0.85\,\mathrm{eV} \] \(\textbf{Distinguish the three energy quantities:}\) The potential energy is negative and twice the total energy. The kinetic energy is positive and equal to the magnitude of the total energy. The ionisation energy is also equal to this magnitude because the electron must be raised from its occupied bound state to the zero-energy reference. These equalities arise from the Coulomb potential and the Bohr energy convention, not from treating potential, kinetic and total energies as interchangeable. \(\textbf{Final answer:}\) \[ \boxed{n=4,\qquad K=0.85\,\mathrm{eV},\qquad E_{\mathrm{ion}}=0.85\,\mathrm{eV}} \]

Explanation: Bohr's atomic model was developed to explain important features of the hydrogen spectrum using discrete electronic energy levels. The predicted energy difference between two permitted states determines the principal photon frequency associated with their transition. Agreement between the calculated and observed principal line wavelength therefore supports the usefulness of energy quantisation. However, an observed spectral line may exhibit finer structure that the elementary Bohr model does not adequately describe. The existence of additional closely spaced components does not imply that atomic energy becomes continuously variable. Instead, the observation indicates that the simplified energy description omits physical details required for the complete spectrum. The success of a model for one measured quantity must be distinguished from its ability to explain every higher-resolution observation. The simple Bohr model also does not account satisfactorily for the relative intensities of spectral lines or their splitting under applied electric and magnetic fields. More complete quantum mechanical treatment is needed to explain these finer details. It would therefore be incorrect to discard the idea of quantised states simply because the elementary model is incomplete. It would be equally incorrect to assume that matching the principal wavelength guarantees a full prediction of every spectral component. The scientifically justified conclusion is that the Bohr model captures important hydrogen-like energy relationships while remaining limited as a complete description of atomic structure and spectra.

Explanation: \(\textbf{Verify the equal-radius condition:}\) For a hydrogen-like one-electron species, the Bohr radius is \[ r_n=a_0\frac{n^2}{Z} \] For hydrogen, \(Z=1\) and \(n=3\). Therefore, \[ r_H=a_0\frac{3^2}{1}=9a_0 \] For \(\mathrm{Be^{3+}}\), \(Z=4\) and \(n=6\). Thus, \[ r_{Be}=a_0\frac{6^2}{4}=9a_0 \] The given equal-radius condition is satisfied. \(\textbf{Compare the electron kinetic energies:}\) For an electron in a Bohr orbit, \[ K_n=13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] For hydrogen, \[ K_H=13.6\frac{1^2}{3^2} \] \[ K_H=\frac{13.6}{9}\,\mathrm{eV} \] For \(\mathrm{Be^{3+}}\), \[ K_{Be}=13.6\frac{4^2}{6^2} \] \[ K_{Be}=13.6\frac{16}{36}\,\mathrm{eV} \] The kinetic-energy ratio is \[ \frac{K_{Be}}{K_H}=\frac{16/36}{1/9} \] \[ \frac{K_{Be}}{K_H}=4 \] \(\textbf{Compare the angular momenta:}\) Bohr's angular-momentum condition is \[ L_n=\frac{nh}{2\pi} \] Hence, \[ \frac{L_{Be}}{L_H}=\frac{6}{3}=2 \] \(\textbf{Interpret the coexistence of equal radii and unequal energies:}\) The two electrons occupy orbits of equal radii but experience different nuclear attractions. The electron in \(\mathrm{Be^{3+}}\) experiences the attraction of a nucleus with four times the proton number. Consequently, its kinetic energy is greater despite its orbital radius being equal to that in hydrogen. Equal orbital radius does not imply equal orbital speed, kinetic energy or angular momentum. The angular momentum is fixed by the principal quantum number, while the energy depends on both \(Z\) and \(n\). \(\textbf{Final answer:}\) \[ \boxed{\frac{K_{Be}}{K_H}=4,\qquad\frac{L_{Be}}{L_H}=2} \]

Explanation: \(\textbf{Relate the revolution period to the Bohr quantum number:}\) The time period of electron revolution is \[ T_n=\frac{2\pi r_n}{v_n} \] For hydrogen in Bohr's model, \[ r_n\propto n^2 \] and \[ v_n\propto\frac{1}{n} \] Combining these relations, \[ T_n\propto n^3 \] \(\textbf{Infer the final principal quantum number:}\) The time period becomes eight times its original value: \[ \frac{T_f}{T_i}=8 \] Therefore, \[ \left(\frac{n_f}{n_i}\right)^3=8 \] The initial state is \(n_i=3\). Substituting, \[ \left(\frac{n_f}{3}\right)^3=8 \] Taking the cube root, \[ \frac{n_f}{3}=2 \] \[ n_f=6 \] \(\textbf{Calculate the initial electronic energy:}\) \[ E_3=-\frac{13.6}{3^2} \] \[ E_3=-\frac{13.6}{9}\approx-1.511\,\mathrm{eV} \] \(\textbf{Calculate the final electronic energy:}\) \[ E_6=-\frac{13.6}{6^2} \] \[ E_6=-\frac{13.6}{36}\approx-0.378\,\mathrm{eV} \] \(\textbf{Determine the absorbed photon energy:}\) Since the atom moves to a higher electronic state, it absorbs energy. \[ E_{\mathrm{photon}}=E_f-E_i \] \[ E_{\mathrm{photon}}=-\frac{13.6}{36}-\left(-\frac{13.6}{9}\right) \] \[ E_{\mathrm{photon}}=13.6\left(\frac19-\frac1{36}\right) \] \[ E_{\mathrm{photon}}=13.6\times\frac3{36} \] \[ E_{\mathrm{photon}}\approx1.133\,\mathrm{eV} \] \(\textbf{Physical interpretation:}\) The orbital revolution period increases because the electron occupies a larger orbit and has a lower orbital speed. However, the frequency of the absorbed radiation is determined by the energy difference between the stationary states. It must not be identified directly with the electron's revolution frequency. The positive photon energy agrees with the upward transition from \(n=3\) to \(n=6\). The observed change in revolution period therefore permits an indirect determination of the electronic excitation energy. \(\textbf{Final answer:}\) \[ \boxed{n_f=6,\qquad E_{\mathrm{photon}}\approx1.133\,\mathrm{eV}} \]

Explanation: \(\textbf{Check the two independent Bohr relations:}\) For a hydrogen-like species, \[ r_n=a_0\frac{n^2}{Z} \] and \[ E_n=-13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] A proposed measurement must satisfy both equations. \(\textbf{Examine case P:}\) Hydrogen has \(Z=1\). For \(n=2\), \[ r_P=a_0\frac{2^2}{1}=4a_0 \] \[ E_P=-13.6\frac{1^2}{2^2}=-3.40\,\mathrm{eV} \] Both values agree with the table. \(\textbf{Examine case Q:}\) For \(\mathrm{He^+}\), \(Z=2\). At \(n=4\), \[ r_Q=a_0\frac{4^2}{2} \] \[ r_Q=8a_0 \] However, the table lists \(4a_0\). Its energy prediction is \[ E_Q=-13.6\frac{2^2}{4^2} \] \[ E_Q=-3.40\,\mathrm{eV} \] The listed energy is correct, but the radius is not. Thus, Q is inconsistent. \(\textbf{Examine case R:}\) For \(\mathrm{Li^{2+}}\), \(Z=3\), \(n=3\). \[ r_R=a_0\frac9{3}=3a_0 \] \[ E_R=-13.6\frac9{9}=-13.60\,\mathrm{eV} \] Both listed values are correct. \(\textbf{Examine case S:}\) For \(\mathrm{Be^{3+}}\), \(Z=4\), \(n=4\). \[ r_S=a_0\frac{16}{4}=4a_0 \] \[ E_S=-13.6\frac{16}{16}=-13.60\,\mathrm{eV} \] Both listed values are also correct. \(\textbf{Why checking only one quantity is insufficient:}\) Cases P and Q have the same total electronic energy despite belonging to different species and different levels. This does not imply that their orbital radii are equal. Energy depends on \(Z^2/n^2\), whereas radius depends on \(n^2/Z\). The row-selection decision must therefore be based on compatibility with both relations rather than agreement with the energy formula alone.

Explanation: \(\textbf{Infer the occupied principal quantum number:}\) The angular momentum permitted by Bohr's model is \[ L_n=\frac{nh}{2\pi} \] The supplied angular momentum is \[ L=\frac{5h}{2\pi} \] Therefore, \[ n=5 \] \(\textbf{Use the orbital radius to determine nuclear charge:}\) For a hydrogen-like species, \[ r_n=a_0\frac{n^2}{Z} \] The measured orbital radius is \[ r_n=6.25a_0 \] Substituting \(n=5\), \[ 6.25a_0=a_0\frac{25}{Z} \] Cancel the common radius constant: \[ 6.25=\frac{25}{Z} \] Therefore, \[ Z=\frac{25}{6.25} \] \[ Z=4 \] \(\textbf{Identify the one-electron species:}\) Atomic number four corresponds to beryllium. A beryllium nucleus has four protons. To retain exactly one electron, the species must be \(\mathrm{Be^{3+}}\). \(\textbf{Determine the relative electron speed:}\) Bohr's velocity relation is \[ v_n\propto\frac{Z}{n} \] For the identified state, \[ v_{\mathrm{Be^{3+}},5}\propto\frac45 \] For ground-state hydrogen, \[ v_{\mathrm{H},1}\propto\frac11 \] Therefore, \[ \frac{v_{\mathrm{Be^{3+}},5}}{v_{\mathrm{H},1}}=\frac45 \] \(\textbf{Physical interpretation:}\) The angular-momentum observation identifies the principal quantum number without requiring knowledge of the nuclear charge. The radius measurement then supplies the independent information needed to determine \(Z\). Although \(\mathrm{Be^{3+}}\) has a stronger nuclear attraction than hydrogen, its electron occupies the fifth orbit rather than the ground state. The combined dependence \(Z/n\) therefore gives a speed smaller than that of the ground-state hydrogen electron. A larger nuclear charge alone is not sufficient to determine the speed comparison. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{Be^{3+}},\qquad\frac{v_{\mathrm{Be^{3+}},5}}{v_{\mathrm{H},1}}=\frac45} \]

Explanation: \(\textbf{Express the energy released during emission:}\) For hydrogen, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] The emitted photon corresponds to the transition \(n_i\rightarrow2\). Its energy is \[ E_{\mathrm{photon}}=13.6\left(\frac14-\frac1{n_i^2}\right)\mathrm{eV} \] \(\textbf{Express ionisation energy from the initial level:}\) Ionisation from the occupied initial state requires raising the electron to the zero-energy limit. Therefore, \[ E_{\mathrm{ion},i}=\frac{13.6}{n_i^2}\,\mathrm{eV} \] This is not the ground-state ionisation energy unless \(n_i=1\). \(\textbf{Apply the measured energy ratio:}\) The emitted photon energy is eight times the initial-state ionisation energy: \[ E_{\mathrm{photon}}=8E_{\mathrm{ion},i} \] Substituting, \[ 13.6\left(\frac14-\frac1{n_i^2}\right)=8\left(\frac{13.6}{n_i^2}\right) \] Cancel \(13.6\): \[ \frac14-\frac1{n_i^2}=\frac8{n_i^2} \] Therefore, \[ \frac14=\frac9{n_i^2} \] \[ n_i^2=36 \] \[ n_i=6 \] \(\textbf{Calculate the photon energy:}\) The transition is \(6\rightarrow2\). \[ E_{\mathrm{photon}}=13.6\left(\frac14-\frac1{36}\right) \] \[ E_{\mathrm{photon}}=13.6\times\frac8{36} \] \[ E_{\mathrm{photon}}\approx3.022\,\mathrm{eV} \] \(\textbf{Determine the wavelength:}\) \[ \lambda=\frac{hc}{E_{\mathrm{photon}}} \] \[ \lambda=\frac{1240}{3.022} \] \[ \lambda\approx410.3\,\mathrm{nm} \] \(\textbf{Check the original condition:}\) The ionisation energy from \(n=6\) is \[ E_{\mathrm{ion},6}=\frac{13.6}{36}\approx0.378\,\mathrm{eV} \] The ratio is \[ \frac{E_{\mathrm{photon}}}{E_{\mathrm{ion},6}}=\frac{3.022}{0.378}\approx8 \] The photon belongs to the Balmer series because the final level is \(n=2\). The reasoning requires distinguishing the energy needed to ionise the original excited state from the much larger ground-state ionisation energy. \(\textbf{Final answer:}\) \[ \boxed{n_i=6,\qquad\lambda\approx410\,\mathrm{nm}} \]

Explanation: \(\textbf{Relate orbital revolution frequency to electron motion:}\) For a circular Bohr orbit, \[ f_{\mathrm{orb}}=\frac{v}{2\pi r} \] Bohr's angular-momentum quantisation gives \[ m_evr=\frac{nh}{2\pi} \] For the first orbit, \(n=1\): \[ m_evr=\frac{h}{2\pi} \] Therefore, \[ r=\frac{h}{2\pi m_ev} \] \(\textbf{Substitute the orbital radius:}\) \[ f_{\mathrm{orb},1}=\frac{v}{2\pi[h/(2\pi m_ev)]} \] Simplifying, \[ f_{\mathrm{orb},1}=\frac{m_ev^2}{h} \] The electron kinetic energy is \[ K_1=\frac12m_ev^2 \] Thus, \[ m_ev^2=2K_1 \] Consequently, \[ f_{\mathrm{orb},1}=\frac{2K_1}{h} \] \(\textbf{Determine the emitted photon frequency:}\) The electronic energies in hydrogen are \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] The photon energy for the \(2\rightarrow1\) transition is \[ E_{\gamma}=13.6\left(1-\frac14\right) \] \[ E_{\gamma}=\frac34(13.6\,\mathrm{eV}) \] Since \(K_1=13.6\,\mathrm{eV}\), \[ E_{\gamma}=\frac34K_1 \] Using Planck's relation, \[ \nu_{21}=\frac{E_{\gamma}}{h} \] \[ \nu_{21}=\frac{3K_1}{4h} \] \(\textbf{Compare the two frequencies:}\) \[ \frac{\nu_{21}}{f_{\mathrm{orb},1}}=\frac{3K_1/(4h)}{2K_1/h} \] Cancel the common factors: \[ \frac{\nu_{21}}{f_{\mathrm{orb},1}}=\frac38 \] \(\textbf{Physical interpretation:}\) The frequency of radiation emitted during a transition is determined by the energy difference between two stationary states. The orbital revolution frequency describes the electron's motion within one Bohr orbit. These frequencies are therefore not generally equal. Bohr's radiation postulate differs from the classical assumption that orbital revolution directly determines the emitted radiation frequency. The numerical comparison makes that distinction explicit within the mathematical structure of the model. \(\textbf{Final answer:}\) \[ \boxed{\frac{\nu_{21}}{f_{\mathrm{orb},1}}=\frac38} \]

Explanation: \(\textbf{Express the first excitation energy:}\) For hydrogen, the energy of the \(n\)th Bohr state is \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] The next higher bound state has principal quantum number \(n+1\). Therefore, \[ E_{\mathrm{first}}=E_{n+1}-E_n \] Substituting the energy formula, \[ E_{\mathrm{first}}=13.6\left(\frac1{n^2}-\frac1{(n+1)^2}\right)\mathrm{eV} \] \(\textbf{Express the initial-state ionisation energy:}\) Ionisation raises the electron from the occupied level to the zero-energy reference. Hence, \[ E_{\mathrm{ion}}=\frac{13.6}{n^2}\,\mathrm{eV} \] \(\textbf{Form the measured energy ratio:}\) \[ \frac{E_{\mathrm{first}}}{E_{\mathrm{ion}}}=\frac{13.6[1/n^2-1/(n+1)^2]}{13.6/n^2} \] Cancelling the common factor, \[ \frac{E_{\mathrm{first}}}{E_{\mathrm{ion}}}=1-\frac{n^2}{(n+1)^2} \] Simplifying, \[ \frac{E_{\mathrm{first}}}{E_{\mathrm{ion}}}=\frac{2n+1}{(n+1)^2} \] \(\textbf{Identify the permitted integer value of }n\textbf{:}\) The measured ratio is \(9/25\). For \(n=4\), \[ \frac{2n+1}{(n+1)^2}=\frac{2(4)+1}{(4+1)^2} \] \[ =\frac9{25} \] Thus, \[ n=4 \] \(\textbf{Calculate the ionisation energy:}\) \[ E_{\mathrm{ion}}=\frac{13.6}{4^2} \] \[ E_{\mathrm{ion}}=\frac{13.6}{16} \] \[ E_{\mathrm{ion}}=0.850\,\mathrm{eV} \] \(\textbf{Verify the excitation fraction:}\) The energy required for \(4\rightarrow5\) is \[ E_{\mathrm{first}}=13.6\left(\frac1{16}-\frac1{25}\right) \] \[ E_{\mathrm{first}}=13.6\times\frac9{400} \] \[ E_{\mathrm{first}}=0.306\,\mathrm{eV} \] Therefore, \[ \frac{0.306}{0.850}=0.36=\frac9{25} \] \(\textbf{Physical interpretation:}\) The first bound-state excitation requires less energy than ionisation because the electron remains attached to the nucleus after excitation. For higher occupied states, the adjacent energy-level spacing becomes smaller relative to the total binding energy. The given ratio allows the occupied level to be identified without any direct measurement of orbital radius. \(\textbf{Final answer:}\) \[ \boxed{n=4,\qquad E_{\mathrm{ion}}=0.850\,\mathrm{eV}} \]

Explanation: \(\textbf{Use angular momentum to identify the final orbit:}\) Bohr's angular-momentum quantisation condition is \[ L_n=\frac{nh}{2\pi} \] For the same electron, \[ \frac{L_f}{L_i}=\frac{n_f}{n_i} \] The angular momentum becomes two-thirds of its original value: \[ \frac{n_f}{3}=\frac23 \] Therefore, \[ n_f=2 \] The transition is \[ 3\rightarrow2 \] \(\textbf{Compare the orbital radii:}\) For hydrogen, \[ r_n=a_0n^2 \] Hence, \[ \frac{r_f}{r_i}=\frac{2^2}{3^2} \] \[ \frac{r_f}{r_i}=\frac49 \] \(\textbf{Determine the electrostatic-force ratio:}\) The attractive force between the nucleus and electron is \[ F=\frac{e^2}{4\pi\varepsilon_0r^2} \] Therefore, its magnitude varies inversely with the square of orbital radius. \[ \frac{F_f}{F_i}=\left(\frac{r_i}{r_f}\right)^2 \] \[ \frac{F_f}{F_i}=\left(\frac94\right)^2 \] \[ \frac{F_f}{F_i}=\frac{81}{16} \] \(\textbf{Calculate the photon energy:}\) For hydrogen, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] The emitted photon carries \[ E_{\mathrm{photon}}=E_i-E_f \] \[ E_{\mathrm{photon}}=13.6\left(\frac14-\frac19\right) \] \[ E_{\mathrm{photon}}=13.6\times\frac5{36} \] \[ E_{\mathrm{photon}}\approx1.889\,\mathrm{eV} \] \(\textbf{Physical interpretation:}\) The electron moves to a smaller orbit and experiences a stronger electrostatic attraction. Its angular momentum decreases because the final principal quantum number is smaller. Its total electronic energy also decreases, so the atom releases a photon. The increase in attractive force and the emission of radiation are related consequences of the downward transition, but they are calculated from different physical relations. The emitted photon belongs to the Balmer series because the transition ends at \(n=2\). \(\textbf{Final answer:}\) \[ \boxed{\frac{F_f}{F_i}=\frac{81}{16},\qquad E_{\mathrm{photon}}\approx1.889\,\mathrm{eV}} \]

Explanation: \(\textbf{Connect ionisation energy with nuclear charge:}\) For a hydrogen-like species, \[ E_n=-13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] The ionisation energy from an occupied level is the magnitude of its negative total electronic energy: \[ E_{\mathrm{ion},n}=13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] The electron occupies \(n=3\), and the measured ionisation energy is \(54.4\,\mathrm{eV}\). \(\textbf{Solve for the nuclear charge:}\) \[ 54.4=13.6\frac{Z^2}{3^2} \] \[ 54.4=13.6\frac{Z^2}{9} \] Rearranging, \[ Z^2=\frac{54.4\times9}{13.6} \] \[ Z^2=36 \] \[ Z=6 \] \(\textbf{Identify the one-electron ion:}\) Atomic number six corresponds to carbon. A neutral carbon atom contains six electrons. To produce a one-electron carbon species, five electrons must be removed. Therefore, the ion is \(\mathrm{C^{5+}}\). \(\textbf{Calculate the Bohr radius:}\) For a hydrogen-like species, \[ r_n=a_0\frac{n^2}{Z} \] Substituting the determined nuclear charge, \[ r_3=52.9\times\frac{3^2}{6}\,\mathrm{pm} \] \[ r_3=52.9\times\frac96\,\mathrm{pm} \] \[ r_3=52.9\times1.5\,\mathrm{pm} \] \[ r_3=79.35\,\mathrm{pm} \] \(\textbf{Verify the inferred state:}\) The total electronic energy of the state is \[ E_3=-13.6\frac{6^2}{3^2} \] \[ E_3=-54.4\,\mathrm{eV} \] The energy required to raise the electron from this state to the zero-energy ionisation limit is therefore \(54.4\,\mathrm{eV}\), matching the observation. The relatively small orbital radius is consistent with the stronger nuclear attraction exerted by a nucleus containing six protons. The result illustrates that ionisation energy from an excited state can identify nuclear charge when the principal quantum number is independently known. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C^{5+}},\qquad r_3\approx79.4\,\mathrm{pm}} \]

Explanation: \(\textbf{Connect the measured force with Coulomb potential energy:}\) For an electron attracted to a nucleus, the electrostatic potential energy is negative. For the Coulomb interaction, \[ U=-\frac{e^2}{4\pi\varepsilon_0r} \] The magnitude of the electrostatic force is \[ F=\frac{e^2}{4\pi\varepsilon_0r^2} \] Multiplying force by orbital radius gives \[ Fr=\frac{e^2}{4\pi\varepsilon_0r} \] Therefore, \[ U=-Fr \] \(\textbf{Relate potential energy to total electronic energy:}\) For a circular Coulomb-bound Bohr orbit, \[ U=-2K \] The total electronic energy is \[ E=K+U \] Substituting \(K=-U/2\), \[ E=-\frac{U}{2}+U \] \[ E=\frac{U}{2} \] Hence, \[ E=-\frac{Fr}{2} \] \(\textbf{Calculate the magnitude of the binding energy:}\) \[ E_{\mathrm{ion}}=-E=\frac{Fr}{2} \] Substituting, \[ E_{\mathrm{ion}}=\frac{(5.15\times10^{-9})(2.12\times10^{-10})}{2} \] \[ E_{\mathrm{ion}}=5.459\times10^{-19}\,\mathrm{J} \] \(\textbf{Convert the result into electronvolts:}\) \[ E_{\mathrm{ion}}=\frac{5.459\times10^{-19}}{1.60\times10^{-19}}\,\mathrm{eV} \] \[ E_{\mathrm{ion}}\approx3.41\,\mathrm{eV} \] \(\textbf{Identify the principal quantum number independently:}\) For hydrogen, \[ r_n=a_0n^2 \] The first Bohr radius is \[ a_0=52.9\,\mathrm{pm}=5.29\times10^{-11}\,\mathrm{m} \] Therefore, \[ n^2=\frac{2.12\times10^{-10}}{5.29\times10^{-11}} \] \[ n^2\approx4 \] \[ n=2 \] \(\textbf{Check consistency with the hydrogen energy formula:}\) For \(n=2\), \[ E_{\mathrm{ion},2}=\frac{13.6}{2^2} \] \[ E_{\mathrm{ion},2}=3.40\,\mathrm{eV} \] The small difference from \(3.41\,\mathrm{eV}\) results from rounding the measured force and radius. The negative total electronic energy is not equal in magnitude to the full product \(Fr\); it equals half that product. The independent radius calculation verifies the identity of the occupied Bohr state. \(\textbf{Final answer:}\) \[ \boxed{E_{\mathrm{ion}}\approx3.41\,\mathrm{eV},\qquad n=2} \]

Explanation: \(\textbf{Calculate the energy of the emitted photon:}\) The photon originates from the hydrogen transition \(4\rightarrow2\). Its energy is \[ E_{\gamma}=13.6\left(\frac1{2^2}-\frac1{4^2}\right)\mathrm{eV} \] \[ E_{\gamma}=13.6\left(\frac14-\frac1{16}\right) \] \[ E_{\gamma}=13.6\times\frac3{16} \] \[ E_{\gamma}=2.55\,\mathrm{eV} \] \(\textbf{Examine an atom initially in }n=1\textbf{:}\) The minimum energy needed for bound-state excitation from \(n=1\) is \[ E_{1\rightarrow2}=10.20\,\mathrm{eV} \] The photon energy of \(2.55\,\mathrm{eV}\) is insufficient. It is also smaller than the ground-state ionisation energy of \(13.6\,\mathrm{eV}\). Therefore, the photon cannot produce the specified electronic excitation or ionisation from \(n=1\). \(\textbf{Examine an atom initially in }n=2\textbf{:}\) The photon energy is exactly equal to the difference between the \(n=2\) and \(n=4\) energies. Therefore, the reverse transition \[ 2\rightarrow4 \] is an allowed bound-state absorption process. However, the ionisation energy from \(n=2\) is \[ E_{\mathrm{ion},2}=\frac{13.6}{4}=3.40\,\mathrm{eV} \] Since \(2.55\,\mathrm{eV}\) is below this threshold, the atom is not ionised. \(\textbf{Examine an atom initially in }n=3\textbf{:}\) The ionisation energy is \[ E_{\mathrm{ion},3}=\frac{13.6}{9} \] \[ E_{\mathrm{ion},3}\approx1.511\,\mathrm{eV} \] Because the photon energy exceeds this value, ionisation is possible. The released electron carries approximately \[ K_3=2.55-1.511\approx1.039\,\mathrm{eV} \] \(\textbf{Examine an atom initially in }n=4\textbf{:}\) The ionisation energy is \[ E_{\mathrm{ion},4}=\frac{13.6}{16}=0.85\,\mathrm{eV} \] The photon can therefore ionise this atom as well. Its released electron carries \[ K_4=2.55-0.85=1.70\,\mathrm{eV} \] \(\textbf{Interpret the different outcomes:}\) The photon has exactly the energy needed for bound-state excitation from \(n=2\) to \(n=4\). The same photon can ionise atoms already occupying the more weakly bound \(n=3\) or \(n=4\) states. Thus, a photon's ability to cause excitation or ionisation depends on the atom's initial state. The processes must not be classified solely by the photon energy without considering the binding energy of the electron receiving it.