Atoms and Nuclei MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Identify the spectral series:}\) Photon P is produced by a transition terminating at \(n_f=3\), so it belongs to the Paschen series. Photon Q is produced by a transition terminating at \(n_f=2\), so it belongs to the Balmer series. \(\textbf{Stage 1 — Calculate the energy of photon P:}\) \[ E_P=13.6\left(\frac{1}{3^2}-\frac{1}{6^2}\right) \] \[ E_P=13.6\left(\frac{1}{9}-\frac{1}{36}\right) \] \[ E_P=13.6\times\frac{3}{36} \] \[ E_P=\frac{13.6}{12}\,\text{eV} \] \[ E_P\approx1.133\,\text{eV} \] \(\textbf{Stage 2 — Calculate the energy of photon Q:}\) \[ E_Q=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \] \[ E_Q=13.6\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ E_Q=13.6\times\frac{5}{36} \] \[ E_Q\approx1.889\,\text{eV} \] \(\textbf{Stage 3 — Compare photon energies:}\) The common factor \(13.6\,\text{eV}\) cancels. \[ \frac{E_P}{E_Q}=\frac{3/36}{5/36} \] \[ \frac{E_P}{E_Q}=\frac{3}{5} \] \(\textbf{Stage 4 — Compare wavelengths:}\) For photons: \[ E_\gamma=\frac{hc}{\lambda} \] Therefore: \[ \frac{\lambda_P}{\lambda_Q}=\frac{E_Q}{E_P} \] \[ \frac{\lambda_P}{\lambda_Q}=\frac{5}{3} \] \(\textbf{Physical check:}\) The Paschen transition produces a smaller photon energy than the specified Balmer transition. Its wavelength must therefore be longer. This agrees with the Paschen series lying in the infrared region, whereas the first Balmer line lies in the visible region. The wavelength ratio is the reciprocal of the photon-energy ratio, not the same ratio. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{E_P}{E_Q},\frac{\lambda_P}{\lambda_Q}\right)=\left(\frac{3}{5},\frac{5}{3}\right)} \]

Explanation: \(\textbf{Identify the series limits:}\) For hydrogen emission: \[ \frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \] The shortest wavelength within a fixed spectral series occurs when the emitted photon energy is greatest. This corresponds to: \[ n_i\to\infty \] Therefore: \[ \frac{1}{\lambda_{\text{limit}}}=\frac{R_H}{n_f^2} \] \[ \lambda_{\text{limit}}=\frac{n_f^2}{R_H} \] \(\textbf{Stage 1 — Lyman limit:}\) For Lyman: \[ n_f=1 \] \[ \lambda_L=\frac{1}{R_H} \] \(\textbf{Stage 2 — Balmer limit:}\) For Balmer: \[ n_f=2 \] \[ \lambda_B=\frac{4}{R_H} \] \(\textbf{Stage 3 — Paschen limit:}\) For Paschen: \[ n_f=3 \] \[ \lambda_P=\frac{9}{R_H} \] Consequently: \[ \lambda_L:\lambda_B:\lambda_P=1:4:9 \] \(\textbf{Stage 4 — Connect the Paschen limit with ionisation:}\) The Paschen series terminates at \(n=3\). Its series-limit photon energy equals the energy difference between \(n=\infty\) and \(n=3\). The energy at the ionisation limit is zero. Therefore: \[ I_3=E_\infty-E_3 \] \[ I_3=0-\left(-\frac{13.6}{9}\right) \] \[ I_3=\frac{13.6}{9}\,\text{eV} \] \(\textbf{Physical interpretation:}\) A series limit corresponds to the largest possible bound-bound transition energy approaching the continuum. The same energy represents the threshold for ionising hydrogen from the terminal level of that series. Thus, the Paschen-series limit is directly related to ionisation from \(n=3\). The wavelength ratios and ionisation threshold follow from the same energy-level structure. \(\textbf{Final answer:}\) \[ \boxed{\left(\lambda_L:\lambda_B:\lambda_P,\ I_3\right)=\left(1:4:9,\frac{13.6}{9}\,\text{eV}\right)} \]

Explanation: \(\textbf{Identify the one-electron systems:}\) Hydrogen has nuclear charge \(Z=1\). Singly ionised helium has nuclear charge \(Z=2\). The energy levels of a hydrogen-like one-electron system are: \[ E_n=-\frac{13.6Z^2}{n^2}\,\text{eV} \] \(\textbf{Stage 1 — Photon energy for hydrogen:}\) For \(n=2\to n=1\): \[ E_{\gamma P}=13.6\left(1-\frac{1}{4}\right) \] \[ E_{\gamma P}=13.6\times\frac{3}{4} \] \[ E_{\gamma P}=10.2\,\text{eV} \] \(\textbf{Stage 2 — Photon energy for the helium ion:}\) For \(n=4\to n=2\), with \(Z=2\): \[ E_{\gamma Q}=13.6(2^2)\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \] \[ E_{\gamma Q}=54.4\left(\frac{1}{4}-\frac{1}{16}\right) \] \[ E_{\gamma Q}=54.4\times\frac{3}{16} \] \[ E_{\gamma Q}=10.2\,\text{eV} \] Thus, both photons have equal energies and equal wavelengths. \(\textbf{Stage 3 — Compare the initial orbital radii:}\) For a hydrogen-like Bohr orbit: \[ r_n=\frac{n^2a_0}{Z} \] The initial hydrogen orbit is \(n=2\): \[ r_{P,i}=\frac{2^2a_0}{1}=4a_0 \] The initial helium-ion orbit is \(n=4\): \[ r_{Q,i}=\frac{4^2a_0}{2}=8a_0 \] Therefore: \[ \frac{r_{Q,i}}{r_{P,i}}=\frac{8a_0}{4a_0} \] \[ \frac{r_{Q,i}}{r_{P,i}}=2 \] \(\textbf{Physical interpretation:}\) Equal photon energies do not require the initial or final Bohr radii to be identical. The greater nuclear charge in the helium ion is compensated by the different principal quantum numbers in the transition-energy calculation. However, radius depends on \(n^2/Z\), so the corresponding initial orbit sizes differ. This comparison demonstrates why spectral equality and orbital-size equality are distinct conditions. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{E_{\gamma Q}}{E_{\gamma P}},\frac{r_{Q,i}}{r_{P,i}}\right)=(1,2)} \]

Explanation: \(\textbf{Identify the two emission paths:}\) The first atom emits two photons: \[ 4\to3\to2 \] The second atom emits one photon: \[ 4\to2 \] Both atoms begin and end at the same energy levels. Therefore, the total energy emitted in the two-step process equals the energy of the single photon emitted in the direct transition. \(\textbf{Stage 1 — Express photon energies:}\) Photon energy is: \[ E_\gamma=\frac{hc}{\lambda} \] For the two-step process: \[ E_{43}+E_{32}=E_{42} \] Substituting the photon-energy relation: \[ \frac{hc}{\lambda_{43}}+\frac{hc}{\lambda_{32}}=\frac{hc}{\lambda_{42}} \] Cancelling \(hc\): \[ \frac{1}{\lambda_{43}}+\frac{1}{\lambda_{32}}=\frac{1}{\lambda_{42}} \] \(\textbf{Stage 2 — Rearrange for the required wavelength:}\) \[ \frac{1}{\lambda_{42}}=\frac{1}{1875}+\frac{1}{656} \] All wavelengths are expressed in nanometres. Combining the fractions: \[ \frac{1}{\lambda_{42}}=\frac{656+1875}{1875\times656} \] \[ \frac{1}{\lambda_{42}}=\frac{2531}{1{,}230{,}000}\,\text{nm}^{-1} \] Therefore: \[ \lambda_{42}=\frac{1{,}230{,}000}{2531}\,\text{nm} \] \[ \lambda_{42}\approx486\,\text{nm} \] \(\textbf{Physical check:}\) The direct-transition photon carries the sum of the two positive photon energies. It must therefore have greater energy than either individual cascade photon. Its wavelength must be shorter than both \(1875\,\text{nm}\) and \(656\,\text{nm}\). The calculated value satisfies this condition. \(\textbf{Conceptual distinction:}\) Photon energies and frequencies add when the corresponding atomic energy differences add. Wavelengths do not add because photon energy is inversely proportional to wavelength. The correct relation involves reciprocal wavelengths. \(\textbf{Final answer:}\) \[ \boxed{\lambda_{42}\approx486\,\text{nm}} \]

Explanation: Bohr's model assigns definite energies to permitted stationary states of hydrogen. When an electron changes between two such states, the emitted or absorbed photon energy equals the difference between the state energies. Consequently, the model predicts the characteristic frequencies of hydrogen spectral lines through the relation \(h\nu=|\Delta E|\). This successfully accounts for the positions of the principal hydrogen spectral lines. However, knowing the energy of an emitted photon does not determine how many photons of that energy will be emitted in a given time. Spectral-line intensity is associated with the amount of radiation emitted at the corresponding frequency. It therefore depends on the population of the relevant excited states and the occurrence rates of the associated transitions. The elementary Bohr model does not provide a satisfactory quantitative description of these transition probabilities and relative line intensities. Thus, the Assertion is true. The Reason is also true because it distinguishes the energy carried by an individual photon from the total radiation produced through many transitions. This distinction explains why successful prediction of spectral-line positions does not automatically imply successful prediction of line intensities. Therefore, the Reason correctly explains the Assertion.

Explanation: \(\textbf{Initial binding energy:}\) Both atoms begin in the second hydrogen stationary state. \[ E_2=-\frac{13.6}{2^2} \] \[ E_2=-3.40\,\text{eV} \] Therefore, the ionisation threshold is: \[ E_{\text{ion},2}=3.40\,\text{eV} \] \(\textbf{Stage 1 — Calculate photon P energy:}\) \[ E_{\gamma P}=\frac{hc}{\lambda_P} \] Substituting: \[ E_{\gamma P}=\frac{1240\,\text{eV\,nm}}{300\,\text{nm}} \] \[ E_{\gamma P}\approx4.133\,\text{eV} \] This exceeds the ionisation threshold. Therefore, ionisation is possible. \(\textbf{Calculate the liberated electron's kinetic energy:}\) \[ K_P=E_{\gamma P}-E_{\text{ion},2} \] \[ K_P=4.133-3.40 \] \[ K_P\approx0.733\,\text{eV} \] \(\textbf{Stage 2 — Calculate photon Q energy:}\) \[ E_{\gamma Q}=\frac{hc}{\lambda_Q} \] \[ E_{\gamma Q}=\frac{1240\,\text{eV\,nm}}{400\,\text{nm}} \] \[ E_{\gamma Q}=3.10\,\text{eV} \] Since: \[ 3.10\,\text{eV}\lt 3.40\,\text{eV} \] photon Q cannot ionise the atom. \(\textbf{Stage 3 — Check the limiting wavelength:}\) The maximum photon wavelength capable of ionising hydrogen from \(n=2\) is: \[ \lambda_{\text{threshold}}=\frac{hc}{E_{\text{ion},2}} \] \[ \lambda_{\text{threshold}}=\frac{1240}{3.40}\,\text{nm} \] \[ \lambda_{\text{threshold}}\approx364.7\,\text{nm} \] A shorter wavelength corresponds to higher photon energy. Therefore, \(300\,\text{nm}\) lies on the ionising side of the threshold, whereas \(400\,\text{nm}\) does not. \(\textbf{Physical interpretation:}\) The relevant threshold depends on the atom's initial energy state. The \(n=2\) electron requires less energy for ionisation than a ground-state hydrogen electron. Once the photon energy exceeds the threshold, the excess can appear as kinetic energy of the liberated electron. \(\textbf{Final answer:}\) \[ \boxed{K_P\approx0.73\,\text{eV}} \] Only photon P can cause ionisation.

Explanation: The spectral series is identified by the final energy level of each emission transition. A Balmer line corresponds to a transition terminating at \(n_f=2\). Since the available initial levels extend up to \(n=6\), the possible Balmer transitions are \(3\to2\), \(4\to2\), \(5\to2\) and \(6\to2\). These produce four distinct Balmer photon energies. A Paschen line corresponds to a transition terminating at \(n_f=3\). The possible Paschen transitions are \(4\to3\), \(5\to3\) and \(6\to3\). These produce three distinct Paschen photon energies. The collection contains many atoms, allowing different atoms to follow different downward pathways. Therefore, all these lines may occur somewhere in the emitted spectrum even though no single atom must undergo every listed transition. A direct transition from \(n=6\) to the ground state would bypass the intermediate states and produce only one photon. Conversely, a cascade passing through several levels produces multiple photons belonging to potentially different series. The count therefore concerns distinct possible spectral lines from the collection, not the number of photons emitted by one atom. This distinction is essential when interpreting a many-atom emission spectrum.

Explanation: \(\textbf{Observed spectral transition:}\) \[ n_i=5,\qquad n_f=2 \] \[ \lambda_{52}=434\,\text{nm} \] \(\textbf{Stage 1 — Express the observed wave number:}\) For hydrogen: \[ \frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \] For \(5\to2\): \[ \frac{1}{\lambda_{52}}=R_H\left(\frac{1}{4}-\frac{1}{25}\right) \] \[ \frac{1}{\lambda_{52}}=R_H\left(\frac{25-4}{100}\right) \] \[ \frac{1}{\lambda_{52}}=\frac{21R_H}{100} \] \(\textbf{Stage 2 — Express the Balmer-series limit:}\) At the series limit: \[ n_i\to\infty \] The final level remains \(n_f=2\). Therefore: \[ \frac{1}{\lambda_{\text{limit}}}=\frac{R_H}{4} \] \(\textbf{Stage 3 — Eliminate the Rydberg constant:}\) Taking the wavelength ratio: \[ \frac{\lambda_{\text{limit}}}{\lambda_{52}}=\frac{21R_H/100}{R_H/4} \] \[ \frac{\lambda_{\text{limit}}}{\lambda_{52}}=\frac{21}{100}\times4 \] \[ \frac{\lambda_{\text{limit}}}{\lambda_{52}}=\frac{84}{100} \] Therefore: \[ \lambda_{\text{limit}}=0.84\times434\,\text{nm} \] \[ \lambda_{\text{limit}}=364.56\,\text{nm} \] \[ \lambda_{\text{limit}}\approx365\,\text{nm} \] \(\textbf{Physical check:}\) The series limit corresponds to the maximum possible photon energy for transitions terminating at \(n=2\). Maximum photon energy implies maximum frequency and minimum wavelength. Therefore, the limit wavelength must be shorter than that of any ordinary bound-bound Balmer line. The calculated wavelength satisfies this requirement. \(\textbf{Final answer:}\) \[ \boxed{\lambda_{\text{limit}}\approx365\,\text{nm}} \]

Explanation: \(\textbf{Hydrogen transition:}\) For hydrogen: \[ Z_H=1 \] The emitted photon energy for \(3\to2\) is: \[ E_{\gamma H}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \] \[ E_{\gamma H}=13.6\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ E_{\gamma H}=13.6\times\frac{5}{36} \] \[ E_{\gamma H}\approx1.889\,\text{eV} \] \(\textbf{Helium-ion transition:}\) For singly ionised helium: \[ Z_{\mathrm{He}}=2 \] The transition is \(6\to4\). Therefore: \[ E_{\gamma\mathrm{He}}=13.6(2^2)\left(\frac{1}{4^2}-\frac{1}{6^2}\right) \] \[ E_{\gamma\mathrm{He}}=54.4\left(\frac{1}{16}-\frac{1}{36}\right) \] \[ E_{\gamma\mathrm{He}}=54.4\left(\frac{36-16}{576}\right) \] \[ E_{\gamma\mathrm{He}}=54.4\times\frac{20}{576} \] \[ E_{\gamma\mathrm{He}}=\frac{17}{9}\,\text{eV} \] \[ E_{\gamma\mathrm{He}}\approx1.889\,\text{eV} \] \(\textbf{Compare photon energies:}\) The two calculated photon energies are equal. \[ E_{\gamma\mathrm{He}}=E_{\gamma H} \] \(\textbf{Compare wavelengths:}\) For photons travelling in vacuum: \[ E_\gamma=\frac{hc}{\lambda} \] Therefore, equal photon energies imply equal frequencies and equal wavelengths. \[ \lambda_{\mathrm{He}}=\lambda_H \] \(\textbf{Physical interpretation:}\) Different one-electron atomic systems can produce photons of the same energy through transitions involving different principal quantum numbers. The stronger nuclear charge in the helium ion increases the energy scale by \(Z^2\). The different initial and final quantum numbers compensate for this factor in the specified transition. This equality does not imply that the two atoms have identical orbital radii or identical stationary-state energies. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_{\gamma\mathrm{He}}}{E_{\gamma H}}=\frac{\lambda_{\mathrm{He}}}{\lambda_H}=1} \]

Explanation: \(\textbf{Initial energy level:}\) The atom begins in \(n=4\). \[ E_4=-\frac{13.6}{4^2} \] \[ E_4=-0.85\,\text{eV} \] \(\textbf{Stage 1 — Total energy emitted during relaxation:}\) The final state is the hydrogen ground state: \[ E_1=-13.6\,\text{eV} \] The total energy emitted equals the decrease in atomic energy. \[ E_{\text{emitted}}=E_4-E_1 \] \[ E_{\text{emitted}}=-0.85-(-13.6) \] \[ E_{\text{emitted}}=12.75\,\text{eV} \] This total is independent of whether the atom returns through one direct transition or several intermediate transitions. \(\textbf{Stage 2 — Ionisation threshold from the initial state:}\) The ionisation limit corresponds to: \[ E_\infty=0 \] The minimum energy required to remove the electron from \(n=4\) is: \[ E_{\text{ion},4}=E_\infty-E_4 \] \[ E_{\text{ion},4}=0-(-0.85) \] \[ E_{\text{ion},4}=0.85\,\text{eV} \] \(\textbf{Stage 3 — Connect the two energy requirements:}\) The total emitted energy and the initial ionisation threshold satisfy: \[ E_{\text{emitted}}+E_{\text{ion},4}=12.75+0.85 \] \[ E_{\text{emitted}}+E_{\text{ion},4}=13.60\,\text{eV} \] This equals the ground-state ionisation energy. \(\textbf{Physical interpretation:}\) An excited electron has more energy than a ground-state electron. It can release substantial energy while returning to the ground state, even though only a small additional energy is required to remove it completely from its initial excited state. The two quantities describe different processes and must not be interchanged. The energy reference at infinity makes their relationship particularly clear. \(\textbf{Final answer:}\) \[ \boxed{(E_{\text{emitted}},E_{\text{ion},4})=(12.75,0.85)\,\text{eV}} \]

Explanation: Statement I is correct because absorption promotes an electron from a lower-energy stationary state to a higher-energy state. The absorbed photon must carry energy equal to the difference between the two allowed atomic energies. Here, both transitions begin at \(n=2\). Statement II is correct because the reverse downward transitions involve exactly the same pairs of stationary states. The photon-energy magnitudes are therefore identical for absorption and emission between each pair. Consequently, the corresponding absorption and emission wavelengths are equal under the same idealised energy-level approximation. Statement III is incorrect because excitation from \(n=2\) to \(n=4\) requires more energy than excitation from \(n=2\) to \(n=3\). Greater photon energy corresponds to shorter wavelength, not longer wavelength. Therefore, the \(2\to4\) absorption line has the shorter wavelength. Statement IV is incorrect because absorption from a particular initial level requires atoms to occupy that level. If all atoms are instead in the ground state, the specified transitions beginning at \(n=2\) cannot occur directly as the same bound-bound absorption processes. The population of initial atomic levels therefore affects the observed absorption spectrum. This explains why emission and absorption can involve the same characteristic wavelengths without requiring both kinds of spectra to display identical line strengths under different preparation conditions.

Explanation: \(\textbf{Identify the detector's selection condition:}\) The detector records only transitions belonging to the Balmer and Paschen spectral series. Therefore, Lyman transitions terminating at \(n=1\) do not contribute to the recorded line count. \(\textbf{Stage 1 — Count the possible Balmer lines:}\) Balmer transitions terminate at: \[ n_f=2 \] With levels available up to \(n=5\), the possible initial levels are: \[ n_i=3,4,5 \] Thus, the possible Balmer transitions are: \[ 3\to2,\qquad4\to2,\qquad5\to2 \] These represent three different energy gaps. Therefore: \[ N_{\text{Balmer}}=3 \] \(\textbf{Stage 2 — Count the possible Paschen lines:}\) Paschen transitions terminate at: \[ n_f=3 \] The possible initial levels are: \[ n_i=4,5 \] Thus: \[ 4\to3,\qquad5\to3 \] These represent two distinct Paschen photon energies. Therefore: \[ N_{\text{Paschen}}=2 \] \(\textbf{Stage 3 — Determine the combined number:}\) The recorded spectral lines belong to two different series. For these specified levels, the five listed photon energies are distinct. Therefore: \[ N_{\text{detected}}=N_{\text{Balmer}}+N_{\text{Paschen}} \] \[ N_{\text{detected}}=3+2 \] \[ N_{\text{detected}}=5 \] \(\textbf{Physical interpretation:}\) The complete set of possible transitions among five levels contains more lines than the detector records. The detector excludes transitions ending at \(n=1\), even though such transitions may occur in the ensemble. Moreover, the five detected lines represent possibilities across many atomic relaxation events. A single atom returning from \(n=5\) to the ground state cannot be assumed to emit all five of these distinct photons. The distinction between possible emitted lines, detected spectral series and photons from one atom must be maintained. \(\textbf{Final answer:}\) \[ \boxed{(N_{\text{Balmer}},N_{\text{Paschen}},N_{\text{detected}})=(3,2,5)} \]

Explanation: \(\textbf{Identify the series condition:}\) The Balmer series consists of hydrogen transitions terminating at: \[ n_f=2 \] The shortest wavelength corresponds to the series limit: \[ n_i\to\infty \] \(\textbf{Stage 1 — Write the series-limit relation:}\) The Rydberg formula is: \[ \frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \] For the Balmer limit: \[ \frac{1}{\lambda_{\text{limit}}}=\frac{R_H}{4} \] Therefore: \[ \lambda_{\text{limit}}=\frac{4}{R_H} \] \(\textbf{Stage 2 — Apply the measured wavelength ratio:}\) For the unknown Balmer line: \[ \frac{1}{\lambda}=R_H\left(\frac{1}{4}-\frac{1}{n_i^2}\right) \] Consequently: \[ \frac{\lambda}{\lambda_{\text{limit}}} =\frac{1/4}{1/4-1/n_i^2} \] The given ratio is: \[ \frac{\lambda}{\lambda_{\text{limit}}}=\frac{9}{8} \] Hence: \[ \frac{9}{8}=\frac{1}{1-4/n_i^2} \] Taking reciprocals: \[ \frac{8}{9}=1-\frac{4}{n_i^2} \] \[ \frac{4}{n_i^2}=\frac{1}{9} \] \[ n_i^2=36 \] \[ n_i=6 \] \(\textbf{Stage 3 — Calculate the wavelength:}\) \[ \lambda=\frac{9}{8}\lambda_{\text{limit}} \] \[ \lambda=\frac{9}{8}\times365\,\text{nm} \] \[ \lambda=410.625\,\text{nm} \] \[ \lambda\approx411\,\text{nm} \] \(\textbf{Physical check:}\) A finite bound-bound Balmer transition must have a wavelength longer than the Balmer-series limit. The calculated wavelength satisfies this requirement. It is also shorter than the first Balmer line because the initial state \(n=6\) lies above \(n=3\). Thus, both the inferred quantum number and the wavelength are consistent with the ordering of the Balmer series. \(\textbf{Final answer:}\) \[ \boxed{n_i=6,\qquad\lambda\approx411\,\text{nm}} \]

Explanation: \(\textbf{Interpret the graph equation:}\) For hydrogen emission from \(n_i\) to \(n_f\): \[ E_\gamma=13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\,\text{eV} \] Introducing: \[ x=\frac{1}{n_i^2} \] gives: \[ E_\gamma=\frac{13.6}{n_f^2}-13.6x \] This is a straight line with negative slope. Its photon-energy intercept is: \[ E_{\text{intercept}}=\frac{13.6}{n_f^2}\,\text{eV} \] \(\textbf{Stage 1 — Identify the final stationary state:}\) The observed intercept is approximately \(1.511\,\text{eV}\). Therefore: \[ 1.511\approx\frac{13.6}{n_f^2} \] \[ n_f^2\approx\frac{13.6}{1.511} \] \[ n_f^2\approx9 \] \[ n_f=3 \] Transitions terminating at \(n_f=3\) belong to the Paschen series. \(\textbf{Stage 2 — Identify the first Paschen transition:}\) The first line of a spectral series is produced by the nearest higher stationary state. Therefore: \[ n_i=4,\qquad n_f=3 \] \(\textbf{Stage 3 — Determine the photon energy:}\) \[ E_{\gamma,43}=13.6\left(\frac{1}{9}-\frac{1}{16}\right) \] \[ E_{\gamma,43}=13.6\left(\frac{7}{144}\right) \] \[ E_{\gamma,43}\approx0.6611\,\text{eV} \] \(\textbf{Stage 4 — Calculate the wavelength:}\) \[ \lambda_{43}=\frac{hc}{E_{\gamma,43}} \] \[ \lambda_{43}=\frac{1240\,\text{eV\,nm}}{0.6611\,\text{eV}} \] \[ \lambda_{43}\approx1876\,\text{nm} \] \(\textbf{Graph-domain check:}\) The physical emission lines require: \[ n_i\gt n_f \] Therefore, the allowed initial quantum numbers are \(4,5,6,\ldots\). The graph intercept corresponds to the limiting case \(n_i\to\infty\), not to the first spectral line. The first Paschen wavelength must be longer than its series-limit wavelength because its photon energy is smaller. The calculated infrared wavelength is consistent with this requirement. \(\textbf{Final answer:}\) \[ \boxed{n_f=3,\qquad\lambda_{\text{first}}\approx1876\,\text{nm}} \]

Explanation: \(\textbf{Distinguish two frequencies:}\) The photon frequency is determined by the energy difference between stationary states. The electron revolution frequency is determined by its orbital speed and circumference. These frequencies are not generally identical. \(\textbf{Stage 1 — Find the emitted photon frequency:}\) The hydrogen energy relation is: \[ E_n=-\frac{13.6}{n^2}\,\text{eV} \] For \(n=3\to n=2\): \[ h\nu_\gamma=E_3-E_2 \] \[ h\nu_\gamma=13.6\left(\frac{1}{4}-\frac{1}{9}\right)\,\text{eV} \] \[ h\nu_\gamma=13.6\times\frac{5}{36}\,\text{eV} \] \(\textbf{Stage 2 — Relate revolution frequency to orbital kinetic energy:}\) The revolution frequency of a circular orbit is: \[ f_n=\frac{v_n}{2\pi r_n} \] Bohr's angular-momentum condition gives: \[ mv_nr_n=\frac{nh}{2\pi} \] Therefore: \[ 2\pi r_n=\frac{nh}{mv_n} \] Substituting into the revolution-frequency expression: \[ f_n=\frac{mv_n^2}{nh} \] Since: \[ K_n=\frac{1}{2}mv_n^2 \] we obtain: \[ hf_n=\frac{2K_n}{n} \] \(\textbf{Stage 3 — Apply the relation to }n=2\textbf{:}\) The kinetic energy in the second orbit is: \[ K_2=\frac{13.6}{2^2}=3.4\,\text{eV} \] Thus: \[ hf_2=\frac{2(3.4)}{2} \] \[ hf_2=3.4\,\text{eV} \] \(\textbf{Stage 4 — Compare the frequencies:}\) \[ \frac{\nu_\gamma}{f_2} =\frac{h\nu_\gamma}{hf_2} \] \[ \frac{\nu_\gamma}{f_2} =\frac{13.6(5/36)}{3.4} \] \[ \frac{\nu_\gamma}{f_2} =\frac{4\times5}{36} \] \[ \frac{\nu_\gamma}{f_2}=\frac{5}{9} \] \(\textbf{Physical interpretation:}\) Bohr's transition rule determines radiation frequency using the separation of two energy levels. The revolution frequency describes motion within one stationary orbit. Their different physical definitions explain why they need not coincide. This distinction is essential for understanding why atomic spectral frequencies cannot simply be identified with classical electron revolution frequencies. \(\textbf{Final answer:}\) \[ \boxed{\frac{\nu_\gamma}{f_2}=\frac{5}{9}} \]

Explanation: \(\textbf{Identify the transition:}\) \[ n_i=5,\qquad n_f=3 \] The electron moves to a lower-energy stationary state and emits radiation. \(\textbf{Stage 1 — Calculate the emitted photon energy:}\) \[ E_\gamma=E_5-E_3 \] Using the hydrogen energy relation: \[ E_\gamma=13.6\left(\frac{1}{3^2}-\frac{1}{5^2}\right) \] \[ E_\gamma=13.6\left(\frac{1}{9}-\frac{1}{25}\right) \] \[ E_\gamma=13.6\left(\frac{25-9}{225}\right) \] \[ E_\gamma=13.6\times\frac{16}{225} \] \[ E_\gamma\approx0.967\,\text{eV} \] \(\textbf{Stage 2 — Compare the Bohr radii:}\) For hydrogen: \[ r_n=n^2a_0 \] Therefore: \[ r_i=25a_0 \] \[ r_f=9a_0 \] \(\textbf{Stage 3 — Compare Coulomb-force magnitudes:}\) The electrostatic-force magnitude is: \[ F=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \] Thus: \[ F\propto\frac{1}{r^2} \] The final-to-initial force ratio is: \[ \frac{F_f}{F_i}=\left(\frac{r_i}{r_f}\right)^2 \] \[ \frac{F_f}{F_i}=\left(\frac{25a_0}{9a_0}\right)^2 \] \[ \frac{F_f}{F_i}=\frac{625}{81} \] \(\textbf{Physical consistency check:}\) The final Bohr orbit has a smaller radius. Therefore, the electron experiences a stronger electrostatic attraction in the final stationary state. Its total mechanical energy becomes more negative, and the energy difference is emitted as a photon. Although the total mechanical energy decreases, the electron's orbital kinetic energy increases in the smaller orbit. These changes are mutually consistent with Coulomb attraction and Bohr's quantisation condition. \(\textbf{Final answer:}\) \[ \boxed{\left(E_\gamma,\frac{F_f}{F_i}\right)=\left(0.967\,\text{eV},\frac{625}{81}\right)} \]

Explanation: \(\textbf{Standing-wave condition:}\) For an allowed circular Bohr orbit, the electron's de Broglie wavelength must fit an integral number of times around the circumference. \[ 2\pi r_n=n\lambda_n \] For hydrogen: \[ r_n=n^2a_0 \] \(\textbf{Stage 1 — Connect wavelength with principal quantum number:}\) Substituting the radius relation: \[ 2\pi n^2a_0=n\lambda_n \] For positive \(n\): \[ \lambda_n=2\pi na_0 \] Therefore: \[ n=\frac{\lambda_n}{2\pi a_0} \] \(\textbf{Use the measured wavelength:}\) \[ n=\frac{1.00}{6.28\times0.053} \] \[ n=\frac{1.00}{0.33284} \] \[ n\approx3.00 \] Since an allowed principal quantum number must be a positive integer: \[ n_i=3 \] \(\textbf{Stage 2 — Calculate the energy released:}\) The direct transition is: \[ n=3\longrightarrow n=1 \] The photon energy is: \[ E_\gamma=13.6\left(1-\frac{1}{9}\right) \] \[ E_\gamma=13.6\times\frac{8}{9} \] \[ E_\gamma\approx12.089\,\text{eV} \] \(\textbf{Stage 3 — Convert photon energy into wavelength:}\) \[ \lambda_\gamma=\frac{hc}{E_\gamma} \] \[ \lambda_\gamma=\frac{1240}{12.089}\,\text{nm} \] \[ \lambda_\gamma\approx102.6\,\text{nm} \] \[ \lambda_\gamma\approx103\,\text{nm} \] \(\textbf{Distinguish the two wavelengths:}\) The measured \(1.00\,\text{nm}\) quantity is the de Broglie wavelength associated with the orbiting electron. The approximately \(103\,\text{nm}\) quantity is the wavelength of the electromagnetic photon emitted during the transition. They correspond to different physical waves and must not be equated. The emitted photon belongs to the Lyman series because the final stationary state is \(n=1\). \(\textbf{Final answer:}\) \[ \boxed{n_i=3,\qquad\lambda_\gamma\approx103\,\text{nm}} \]

Explanation: The four quantities depend differently on the principal quantum number in Bohr's model. For hydrogen, the orbital radius varies as the square of the principal quantum number. Therefore, increasing \(n\) from 2 to 4 increases the radius by a factor of four, making measurement P consistent. The orbital speed varies inversely with \(n\). Thus, the fourth-orbit speed is half the second-orbit speed, making measurement Q consistent. Revolution period is the orbital circumference divided by orbital speed. Since circumference increases fourfold while speed decreases to one-half, the period increases eightfold. Measurement R is therefore also consistent. However, the electrostatic-force magnitude depends on the inverse square of orbital radius. A fourfold increase in radius therefore reduces the force to one-sixteenth of its original value, not one-eighth. Measurement S is inconsistent with the Coulomb-force relation. The incorrect value of \(1/8\) resembles the period scaling factor but does not describe the force. This illustrates why different orbital quantities cannot all be assigned the same power of the principal quantum number. The correct force ratio is: \[ \frac{F_4}{F_2}=\left(\frac{r_2}{r_4}\right)^2=\left(\frac{1}{4}\right)^2=\frac{1}{16} \]

Explanation: Initially, atom A is in the first excited state and atom B is in the ground state. The energy difference between hydrogen's \(n=2\) and \(n=1\) states is \(10.2\,\text{eV}\). When atom A undergoes the downward transition, its internal energy decreases by this amount. The emitted photon carries exactly the lost energy under the stated approximation. The photon then reaches atom B, which is initially in the ground state. Since the photon energy matches the ground-to-first-excited-state energy difference, atom B can absorb it. Atom B consequently moves from \(n=1\) to \(n=2\). Its internal energy increases by \(10.2\,\text{eV}\), exactly compensating for the decrease in atom A. The final arrangement therefore has atom A in the ground state and atom B in the first excited state. The location of the excitation has changed, but the combined internal energy of the two atoms has not. The photon transfers excitation energy from one atom to another rather than creating an additional excitation. The process also demonstrates why the same characteristic photon energy can be associated with emission from one atom and absorption by another. The possibility of absorption depends on atom B initially occupying the appropriate lower energy state.

Explanation: \(\textbf{Identify the fixed initial state:}\) The electron begins in: \[ n_i=5 \] Hydrogen's bound-state energy is: \[ E_n=-\frac{13.6}{n^2}\,\text{eV} \] \(\textbf{Stage 1 — Construct the photon-energy expression:}\) For emission: \[ E_\gamma=E_i-E_f \] Substituting: \[ E_\gamma=-\frac{13.6}{25}+\frac{13.6}{n_f^2} \] Therefore: \[ E_\gamma=13.6\left(\frac{1}{n_f^2}-\frac{1}{25}\right)\,\text{eV} \] \(\textbf{Stage 2 — Express the graph in terms of }x\textbf{:}\) The horizontal variable is: \[ x=\frac{1}{n_f^2} \] Thus: \[ E_\gamma=13.6x-\frac{13.6}{25} \] \[ E_\gamma=13.6x-0.544 \] where energy is expressed in electronvolts. \(\textbf{Determine slope and intercept:}\) The coefficient of \(x\) is positive. Therefore: \[ m=+13.6\,\text{eV} \] The extrapolated vertical-axis intercept is: \[ E_{\text{intercept}}=-0.544\,\text{eV} \] \(\textbf{Stage 3 — Determine the physically allowed domain:}\) Emission requires a downward transition: \[ n_f\lt n_i \] Since \(n_i=5\): \[ n_f=1,2,3,4 \] Therefore: \[ \frac{1}{n_f^2}\gt \frac{1}{25} \] \[ x\gt \frac{1}{25} \] \(\textbf{Physical interpretation:}\) The negative extrapolated intercept does not represent an emitted photon with negative energy. It lies outside the domain of the actual downward transitions. Only the discrete points associated with permitted final quantum numbers correspond to physical emission lines. The positive slope arises because lower final states have larger \(1/n_f^2\) and produce larger photon energies. This graph differs from one in which the final level is fixed and the initial level varies. \(\textbf{Final answer:}\) \[ \boxed{m=+13.6\,\text{eV},\quad E_{\text{intercept}}=-0.544\,\text{eV},\quad x\gt \frac{1}{25}} \]