401. In a Rutherford-scattering comparison, two thin foils have target nuclear charges \(Z\) and \(3Z\). The incident \(\alpha\)-particle kinetic energy is changed from \(K\) to \(2K\), and the same scattering angle is counted. If \(N\propto \frac{Z^2}{K^2}\), then \(\frac{N_2}{N_1}\) is
ⓐ. \(\frac{9}{4}\)
ⓑ. \(\frac{3}{2}\)
ⓒ. \(\frac{4}{9}\)
ⓓ. \(36\)
Correct Answer: \(\frac{9}{4}\)
Explanation: \( \textbf{Given proportionality:} \)
\[
N\propto \frac{Z^2}{K^2}
\]
\( \textbf{Case 1:} \) Nuclear charge factor is \(Z\), kinetic energy is \(K\).
\[
N_1\propto \frac{Z^2}{K^2}
\]
\( \textbf{Case 2:} \) Nuclear charge factor is \(3Z\), kinetic energy is \(2K\).
\[
N_2\propto \frac{(3Z)^2}{(2K)^2}
\]
\( \textbf{Simplification:} \)
\[
N_2\propto \frac{9Z^2}{4K^2}
\]
\( \textbf{Ratio:} \)
\[
\frac{N_2}{N_1}=\frac{\frac{9Z^2}{4K^2}}{\frac{Z^2}{K^2}}=\frac{9}{4}
\]
\( \textbf{Final answer:} \) The scattered count becomes \(\frac{9}{4}\) times the original count.
402. An \(\alpha\)-particle of kinetic energy \(K\) has closest approach \(r_0\) to a nucleus of charge \(+Ze\). Another \(\alpha\)-particle approaches a nucleus of charge \(+4Ze\) with kinetic energy \(2K\). The new closest approach is
ⓐ. \(\frac{r_0}{2}\)
ⓑ. \(r_0\)
ⓒ. \(2r_0\)
ⓓ. \(8r_0\)
Correct Answer: \(2r_0\)
Explanation: \( \textbf{Closest-approach relation:} \)
\[
r_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K}
\]
\( \textbf{Proportionality:} \)
\[
r_0\propto \frac{Z}{K}
\]
\( \textbf{Original case:} \)
\[
r_1\propto \frac{Z}{K}
\]
\( \textbf{New case:} \)
\[
r_2\propto \frac{4Z}{2K}
\]
\( \textbf{Ratio:} \)
\[
\frac{r_2}{r_1}=\frac{\frac{4Z}{2K}}{\frac{Z}{K}}=2
\]
\( \textbf{Physical meaning:} \) The larger nuclear charge repels more strongly, and the doubled kinetic energy only partly compensates for it.
\( \textbf{Final answer:} \) The new closest approach is \(2r_0\).
403. For a hydrogen-like ion, \(r_n=\frac{n^2}{Z}a_0\), \(v_n=\frac{Z}{n}v_{\mathrm{H},1}\), and \(E_n=-13.6\frac{Z^2}{n^2}\,eV\). Compared with hydrogen in \(n=2\), \(Li^{2+}\) in \(n=6\) has
ⓐ. radius \(3\) times larger, speed same, energy \(\frac{1}{9}\) as negative
ⓑ. radius \(9\) times larger, speed half, energy same
ⓒ. radius same, speed \(3\) times larger, energy \(9\) times more negative
ⓓ. radius \(3\) times larger, speed same, energy same
Correct Answer: radius \(3\) times larger, speed same, energy same
Explanation: \( \textbf{Hydrogen case:} \) \(Z_H=1\), \(n_H=2\).
\[
r_H=\frac{2^2}{1}a_0=4a_0
\]
\[
v_H=\frac{1}{2}v_{\mathrm{H},1}
\]
\[
E_H=-13.6\frac{1^2}{2^2}\,eV=-3.4\,eV
\]
\( \textbf{\(Li^{2+}\) case:} \) \(Z=3\), \(n=6\).
\[
r_{Li}=\frac{6^2}{3}a_0=12a_0
\]
\[
v_{Li}=\frac{3}{6}v_{\mathrm{H},1}=\frac{1}{2}v_{\mathrm{H},1}
\]
\[
E_{Li}=-13.6\frac{3^2}{6^2}\,eV=-13.6\frac{9}{36}\,eV=-3.4\,eV
\]
\( \textbf{Comparison:} \) Radius ratio \(=\frac{12a_0}{4a_0}=3\), speed is equal, and energy is equal.
\( \textbf{Final answer:} \) The radius is \(3\) times larger, while speed and energy are the same.
404. During the transition \(n_i=4\to n_f=2\), a one-electron ion emits a photon of wavelength \(30.4\,\text{nm}\). If the corresponding hydrogen transition \(4\to2\) has wavelength \(486\,\text{nm}\), the value of \(Z\) is closest to
ⓐ. \(2\)
ⓑ. \(3\)
ⓒ. \(5\)
ⓓ. \(4\)
Correct Answer: \(4\)
Explanation: \( \textbf{Same-transition scaling:} \) For hydrogen-like ions, photon energy for the same transition varies as \(Z^2\).
\( \textbf{Wavelength scaling:} \)
\[
\lambda_Z=\frac{\lambda_H}{Z^2}
\]
\( \textbf{Given values:} \)
\[
30.4=\frac{486}{Z^2}
\]
\( \textbf{Solve for } Z^2\textbf{:} \)
\[
Z^2=\frac{486}{30.4}\approx15.99
\]
\( \textbf{Therefore:} \)
\[
Z\approx4
\]
\( \textbf{Physical interpretation:} \) A \(4\)-times larger nuclear charge number gives \(16\)-times larger transition energy and about \(16\)-times shorter wavelength.
\( \textbf{Final answer:} \) \(Z\approx4\).
405. In hydrogen, a sample is excited up to \(n=5\). Only the spectral lines ending at \(n=2\) are observed. The number of possible Balmer lines from this sample is
ⓐ. \(2\)
ⓑ. \(3\)
ⓒ. \(4\)
ⓓ. \(10\)
Correct Answer: \(3\)
Explanation: \( \textbf{Highest level:} \) The sample has atoms excited up to \(n=5\).
\( \textbf{Balmer condition:} \) Balmer lines end at \(n=2\).
\( \textbf{Allowed upper levels for Balmer lines:} \)
\[
n_i=3,4,5
\]
\( \textbf{Possible transitions:} \)
\[
3\to2,\quad 4\to2,\quad 5\to2
\]
\( \textbf{Count:} \)
\[
N=3
\]
\( \textbf{Contrast:} \) The formula \(\frac{n(n-1)}{2}\) would count all possible lines among levels up to \(n=5\), not only the Balmer subset.
\( \textbf{Final answer:} \) There are \(3\) possible Balmer lines.
406. A hydrogen atom in \(n=3\) absorbs a photon of wavelength \(821\,\text{nm}\). Taking \(hc=1240\,eV\,\text{nm}\), the most suitable outcome is
ⓐ. ionisation, \(K\approx0\)
ⓑ. excitation to \(n=4\)
ⓒ. emission of a \(3\to2\) photon
ⓓ. excitation to \(n=2\)
Correct Answer: ionisation, \(K\approx0\)
Explanation: \( \textbf{Photon energy:} \)
\[
E=\frac{hc}{\lambda}
\]
\( \textbf{Substitution:} \)
\[
E=\frac{1240}{821}\,eV
\]
\( \textbf{Calculation:} \)
\[
E\approx1.51\,eV
\]
\( \textbf{Hydrogen energy at } n=3\textbf{:} \)
\[
E_3=-\frac{13.6}{9}\,eV\approx-1.51\,eV
\]
\( \textbf{Ionisation threshold from } n=3\textbf{:} \)
\[
0-(-1.51)=1.51\,eV
\]
\( \textbf{Outcome:} \) The photon just supplies the energy needed to remove the electron from \(n=3\).
\( \textbf{Final answer:} \) The atom is ionised with nearly zero kinetic energy.
407. A hydrogen atom emits two photons in the cascade \(5\to3\to2\). The total emitted energy is equal to that of the direct transition
ⓐ. \(5\to1\)
ⓑ. \(3\to1\)
ⓒ. \(5\to2\)
ⓓ. \(2\to1\)
Correct Answer: \(5\to2\)
Explanation: \( \textbf{Energy of first photon:} \)
\[
E_{5\to3}=E_5-E_3
\]
\( \textbf{Energy of second photon:} \)
\[
E_{3\to2}=E_3-E_2
\]
\( \textbf{Total emitted energy:} \)
\[
(E_5-E_3)+(E_3-E_2)
\]
\( \textbf{Cancellation:} \)
\[
E_5-E_3+E_3-E_2=E_5-E_2
\]
\( \textbf{Meaning:} \) This is exactly the energy of a direct \(5\to2\) photon.
\( \textbf{Final answer:} \) The total cascade energy equals the direct \(5\to2\) transition energy.
408. For \(Z=2\), a one-electron ion makes a transition from \(n=4\) to \(n=2\). Using the hydrogen \(4\to2\) energy \(2.55\,eV\), the emitted photon energy and wavelength are closest to
ⓐ. \(2.55\,eV\) and \(486\,\text{nm}\)
ⓑ. \(5.10\,eV\) and \(243\,\text{nm}\)
ⓒ. \(40.80\,eV\) and \(30.4\,\text{nm}\)
ⓓ. \(10.20\,eV\) and \(122\,\text{nm}\)
Correct Answer: \(10.20\,eV\) and \(122\,\text{nm}\)
Explanation: \( \textbf{Same-transition energy scaling:} \)
\[
E_Z=Z^2E_H
\]
\( \textbf{For } Z=2\textbf{:} \)
\[
E_Z=2^2(2.55\,eV)
\]
\( \textbf{Energy:} \)
\[
E_Z=10.20\,eV
\]
\( \textbf{Using } hc=1240\,eV\,\text{nm}\textbf{:} \)
\[
\lambda=\frac{1240}{10.20}\,\text{nm}
\]
\( \textbf{Calculation:} \)
\[
\lambda\approx122\,\text{nm}
\]
\( \textbf{Interpretation:} \) The \(He^+\)-like corresponding transition has \(4\) times the energy and one-fourth the wavelength of the hydrogen line.
\( \textbf{Final answer:} \) The photon energy is \(10.20\,eV\), and the wavelength is about \(122\,\text{nm}\).
409. In a Bohr orbit, the electron’s angular momentum is \(5\hbar\), and the orbit radius is \(25a_0\). If the atom is hydrogen, the electron speed is
ⓐ. \(5v_{\mathrm{H},1}\)
ⓑ. \(\frac{v_{\mathrm{H},1}}{5}\)
ⓒ. \(\frac{v_{\mathrm{H},1}}{25}\)
ⓓ. \(25v_{\mathrm{H},1}\)
Correct Answer: \(\frac{v_{\mathrm{H},1}}{5}\)
Explanation: \( \textbf{Angular momentum condition:} \)
\[
L=n\hbar
\]
\( \textbf{Given:} \)
\[
L=5\hbar
\]
\( \textbf{Thus:} \)
\[
n=5
\]
\( \textbf{Speed formula for hydrogen:} \)
\[
v_n=\frac{v_{\mathrm{H},1}}{n}
\]
\( \textbf{Substitution:} \)
\[
v_5=\frac{v_{\mathrm{H},1}}{5}
\]
\( \textbf{Radius check:} \) For hydrogen, \(r_5=5^2a_0=25a_0\), so the given radius is consistent.
\( \textbf{Final answer:} \) The electron speed is \(\frac{v_{\mathrm{H},1}}{5}\).
410. A hydrogen atom in \(n=2\) absorbs a photon of wavelength \(365\,\text{nm}\). Taking \(hc=1240\,eV\,\text{nm}\), the most suitable result is
ⓐ. excitation to \(n=3\)
ⓑ. ionisation, \(K\approx0\)
ⓒ. excitation to \(n=4\)
ⓓ. emission of a Balmer photon
Correct Answer: ionisation, \(K\approx0\)
Explanation: \( \textbf{Photon energy:} \)
\[
E=\frac{hc}{\lambda}
\]
\( \textbf{Substitution:} \)
\[
E=\frac{1240}{365}\,eV
\]
\( \textbf{Calculation:} \)
\[
E\approx3.40\,eV
\]
\( \textbf{Hydrogen energy at } n=2\textbf{:} \)
\[
E_2=-3.40\,eV
\]
\( \textbf{Ionisation energy from } n=2\textbf{:} \)
\[
0-(-3.40)=3.40\,eV
\]
\( \textbf{Conclusion:} \) The photon energy matches the Balmer-limit energy.
\( \textbf{Final answer:} \) The atom is just ionised from \(n=2\).
411. Two hydrogen transitions have wavelengths \(\lambda_P\) and \(\lambda_Q\). Transition P is \(4\to2\), and transition Q is \(4\to1\). The ratio \(\frac{\lambda_P}{\lambda_Q}\) is
ⓐ. \(\frac{1}{5}\)
ⓑ. \(\frac{16}{3}\)
ⓒ. \(\frac{5}{1}\)
ⓓ. \(\frac{3}{16}\)
Correct Answer: \(\frac{5}{1}\)
Explanation: \( \textbf{Photon wavenumber relation:} \)
\[
\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
\]
\( \textbf{For P, } 4\to2\textbf{:} \)
\[
\frac{1}{\lambda_P}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)
\]
\[
\frac{1}{\lambda_P}=R\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3R}{16}
\]
\( \textbf{For Q, } 4\to1\textbf{:} \)
\[
\frac{1}{\lambda_Q}=R\left(1-\frac{1}{16}\right)=\frac{15R}{16}
\]
\( \textbf{Wavelength ratio:} \)
\[
\frac{\lambda_P}{\lambda_Q}=\frac{\frac{1}{\frac{3R}{16}}}{\frac{1}{\frac{15R}{16}}}=\frac{15R/16}{3R/16}=5
\]
\( \textbf{Final answer:} \) \(\lambda_P:\lambda_Q=5:1\).
412. The second Bohr level of a one-electron ion has energy \(E_2=-30.6\,eV\). The value of \(Z\) is
ⓐ. \(2\)
ⓑ. \(3\)
ⓒ. \(4\)
ⓓ. \(6\)
Correct Answer: \(3\)
Explanation: \( \textbf{Energy formula:} \)
\[
E_n=-13.6\frac{Z^2}{n^2}\,eV
\]
\( \textbf{Given:} \)
\[
E_2=-30.6\,eV
\]
\( \textbf{Substitution:} \)
\[
-30.6=-13.6\frac{Z^2}{2^2}
\]
\( \textbf{Remove negative sign:} \)
\[
30.6=13.6\frac{Z^2}{4}
\]
\( \textbf{Solve for } Z^2\textbf{:} \)
\[
Z^2=\frac{30.6\times4}{13.6}=9
\]
\( \textbf{Therefore:} \)
\[
Z=3
\]
\( \textbf{Final answer:} \) The ion has \(Z=3\).
413. A proposed hydrogen transition has photon energy \(5.0\,eV\) and is claimed to be a bound-bound transition from \(n=2\). The claim should be rejected because
ⓐ. all photons from \(n=2\) must have exactly \(10.2\,eV\)
ⓑ. bound-bound absorption from \(n=2\) requires negative photon energy
ⓒ. photons cannot interact with excited hydrogen atoms
ⓓ. \(5.0\,eV\) exceeds the \(n=2\) ionisation energy
Correct Answer: \(5.0\,eV\) exceeds the \(n=2\) ionisation energy
Explanation: The \(n=2\) level of hydrogen has energy \(-3.4\,eV\). The ionisation limit is \(0\,eV\), so only \(3.4\,eV\) is needed to free the electron. A photon of \(5.0\,eV\) has more than enough energy to ionise the atom, leaving excess kinetic energy. Bound-bound absorption must match a specific gap between two negative-energy levels. Since \(5.0\,eV\) exceeds the ionisation threshold from \(n=2\), it cannot correspond to a bound-bound transition starting from \(n=2\).
414. In a one-electron ion, the first orbit radius is \(\frac{a_0}{5}\), and the first orbit speed is
ⓐ. \(\frac{v_{\mathrm{H},1}}{5}\)
ⓑ. \(25v_{\mathrm{H},1}\)
ⓒ. \(5v_{\mathrm{H},1}\)
ⓓ. \(\frac{v_{\mathrm{H},1}}{25}\)
Correct Answer: \(5v_{\mathrm{H},1}\)
Explanation: \( \textbf{First-orbit radius formula:} \)
\[
r_1=\frac{a_0}{Z}
\]
\( \textbf{Given:} \)
\[
r_1=\frac{a_0}{5}
\]
\( \textbf{Therefore:} \)
\[
Z=5
\]
\( \textbf{Speed formula:} \)
\[
v_n=\frac{Z}{n}v_{\mathrm{H},1}
\]
\( \textbf{For the first orbit:} \)
\[
v_1=Zv_{\mathrm{H},1}
\]
\( \textbf{Substitution:} \)
\[
v_1=5v_{\mathrm{H},1}
\]
\( \textbf{Final answer:} \) The first-orbit speed is \(5v_{\mathrm{H},1}\).
415. A graph of \(E_n\) versus \(\frac{1}{n^2}\) for a hydrogen-like ion has slope \(-122.4\,eV\). The value of \(Z\) is
ⓐ. \(3\)
ⓑ. \(2\)
ⓒ. \(4\)
ⓓ. \(9\)
Correct Answer: \(3\)
Explanation: \( \textbf{Energy formula:} \)
\[
E_n=-13.6Z^2\left(\frac{1}{n^2}\right)\,eV
\]
\( \textbf{Graph form:} \) If \(E_n\) is plotted against \(\frac{1}{n^2}\), the slope is
\[
-13.6Z^2\,eV
\]
\( \textbf{Given slope:} \)
\[
-13.6Z^2=-122.4
\]
\( \textbf{Solve:} \)
\[
Z^2=\frac{122.4}{13.6}=9
\]
\[
Z=3
\]
\( \textbf{Interpretation:} \) The graph corresponds to a hydrogen-like ion such as \(Li^{2+}\).
\( \textbf{Final answer:} \) \(Z=3\).
416. In a one-electron ion, an electron in \(n=3\) has the same speed as an electron in hydrogen at \(n=1\). The ion’s \(Z\) is
ⓐ. \(3\)
ⓑ. \(1\)
ⓒ. \(2\)
ⓓ. \(9\)
Correct Answer: \(3\)
Explanation: \( \textbf{Speed formula:} \)
\[
v_n=\frac{Z}{n}v_{\mathrm{H},1}
\]
\( \textbf{Hydrogen at } n=1\textbf{:} \)
\[
v_H=v_{\mathrm{H},1}
\]
\( \textbf{Ion at } n=3\textbf{:} \)
\[
v_{\text{ion}}=\frac{Z}{3}v_{\mathrm{H},1}
\]
\( \textbf{Equal speed condition:} \)
\[
\frac{Z}{3}v_{\mathrm{H},1}=v_{\mathrm{H},1}
\]
\( \textbf{Solve:} \)
\[
Z=3
\]
\( \textbf{Final answer:} \) The ion must have \(Z=3\).
417. A hydrogen atom is excited to \(n=6\). The maximum number of distinct emission lines that end specifically in \(n=3\) is
ⓐ. \(2\)
ⓑ. \(3\)
ⓒ. \(6\)
ⓓ. \(15\)
Correct Answer: \(3\)
Explanation: \( \textbf{Final level condition:} \) The lines must end at \(n=3\), so they belong to the Paschen series.
\( \textbf{Highest level available:} \) \(n=6\).
\( \textbf{Possible upper levels:} \)
\[
n_i=4,5,6
\]
\( \textbf{Possible transitions:} \)
\[
4\to3,\quad 5\to3,\quad 6\to3
\]
\( \textbf{Count:} \)
\[
N=3
\]
\( \textbf{Contrast:} \) The total number of all possible lines from \(n=6\) would be \(\frac{6(5)}{2}=15\), but only \(3\) of them end at \(n=3\).
\( \textbf{Final answer:} \) There are \(3\) such lines.
418. In hydrogen, the frequency of the \(2\to1\) photon is compared with the frequency of the \(4\to2\) photon. The ratio \(\nu_{2\to1}:\nu_{4\to2}\) is
ⓐ. \(1:4\)
ⓑ. \(2:1\)
ⓒ. \(4:1\)
ⓓ. \(5:1\)
Correct Answer: \(4:1\)
Explanation: \( \textbf{Frequency is proportional to photon energy:} \)
\[
E=h\nu
\]
\( \textbf{For } 2\to1\textbf{:} \)
\[
E_{2\to1}=13.6\left(1-\frac{1}{4}\right)\,eV
\]
\[
E_{2\to1}=13.6\left(\frac{3}{4}\right)=10.2\,eV
\]
\( \textbf{For } 4\to2\textbf{:} \)
\[
E_{4\to2}=13.6\left(\frac{1}{4}-\frac{1}{16}\right)\,eV
\]
\[
E_{4\to2}=13.6\left(\frac{3}{16}\right)=2.55\,eV
\]
\( \textbf{Frequency ratio:} \)
\[
\frac{\nu_{2\to1}}{\nu_{4\to2}}=\frac{10.2}{2.55}=4
\]
\( \textbf{Final answer:} \) The ratio is \(4:1\).
419. A one-electron ion shows a transition \(n=3\to n=2\) with photon energy \(17.0\,eV\). Since the hydrogen \(3\to2\) energy is about \(1.89\,eV\), the ion is closest to
ⓐ. \(Li^{2+}\)
ⓑ. \(He^+\)
ⓒ. \(Be^{3+}\)
ⓓ. \(B^{4+}\)
Correct Answer: \(Li^{2+}\)
Explanation: \( \textbf{Same-transition scaling:} \)
\[
E_Z=Z^2E_H
\]
\( \textbf{Given values:} \)
\[
17.0\approx Z^2(1.89)
\]
\( \textbf{Solve for } Z^2\textbf{:} \)
\[
Z^2\approx\frac{17.0}{1.89}\approx9.0
\]
\( \textbf{Therefore:} \)
\[
Z\approx3
\]
\( \textbf{Ion identification:} \) A one-electron ion with \(Z=3\) is \(Li^{2+}\).
\( \textbf{Final answer:} \) The ion is closest to \(Li^{2+}\).
420. In a Bohr-de Broglie orbit, the circumference initially contains \(n\) wavelengths. If the orbit changes to another allowed orbit where the radius becomes \(4\) times larger and the electron speed becomes half, the new number of wavelengths around the circumference becomes
ⓐ. \(4n\)
ⓑ. \(8n\)
ⓒ. \(\frac{n}{2}\)
ⓓ. \(2n\)
Correct Answer: \(2n\)
Explanation: \( \textbf{Initial number of wavelengths:} \)
\[
n=\frac{2\pi r}{\lambda}
\]
\( \textbf{de Broglie relation:} \)
\[
\lambda=\frac{h}{m_ev}
\]
\( \textbf{Speed change:} \) If the electron speed becomes half, the de Broglie wavelength becomes double.
\[
\lambda'=2\lambda
\]
\( \textbf{Radius change:} \)
\[
r'=4r
\]
\( \textbf{New number of wavelengths:} \)
\[
n'=\frac{2\pi r'}{\lambda'}=\frac{2\pi(4r)}{2\lambda}
\]
\( \textbf{Simplification:} \)
\[
n'=2\frac{2\pi r}{\lambda}=2n
\]
\( \textbf{Final answer:} \) The new number of wavelengths is \(2n\).
