301. A hydrogen atom emits a photon of energy \(1.89\,eV\) in a transition. Using \(hc=1240\,eV\,\text{nm}\), the wavelength of the photon is closest to
ⓐ. \(656\,\text{nm}\)
ⓑ. \(122\,\text{nm}\)
ⓒ. \(365\,\text{nm}\)
ⓓ. \(1240\,\text{nm}\)
Correct Answer: \(656\,\text{nm}\)
Explanation: \( \textbf{Given photon energy:} \) \(E=1.89\,eV\).
\( \textbf{Useful relation:} \)
\[
E=\frac{hc}{\lambda}
\]
\( \textbf{Rearranged form:} \)
\[
\lambda=\frac{hc}{E}
\]
\( \textbf{Given constant form:} \) \(hc=1240\,eV\,\text{nm}\).
\( \textbf{Substitution:} \)
\[
\lambda=\frac{1240}{1.89}\,\text{nm}
\]
\( \textbf{Calculation:} \)
\[
\lambda\approx656\,\text{nm}
\]
\( \textbf{Spectral meaning:} \) This lies in the visible red region and matches the first Balmer line of hydrogen.
\( \textbf{Final answer:} \) The wavelength is closest to \(656\,\text{nm}\).
302. In hydrogen, two downward transitions are compared: transition P is \(4\to2\), and transition Q is \(4\to1\). The photon from transition Q has
ⓐ. smaller energy and longer wavelength than P
ⓑ. greater energy and shorter wavelength than P
ⓒ. the same energy as P because both start from \(n=4\)
ⓓ. zero energy because both are downward transitions
Correct Answer: greater energy and shorter wavelength than P
Explanation: The emitted photon energy depends on the difference between the initial and final energy levels. Both transitions start at \(n=4\), but they end at different lower levels. The level \(n=1\) is much lower in energy than \(n=2\), so the drop \(4\to1\) is larger than the drop \(4\to2\). A larger energy drop gives a higher-energy photon. Since \(E=\frac{hc}{\lambda}\), the higher-energy photon has a shorter wavelength. The starting level alone is not enough to decide photon energy; the final level is equally important.
303. A hydrogen atom initially in \(n=2\) absorbs a photon of energy \(5.0\,eV\). Since the ionisation energy from \(n=2\) is \(3.4\,eV\), the kinetic energy of the emitted electron is
ⓐ. \(3.4\,eV\)
ⓑ. \(1.6\,eV\)
ⓒ. \(5.0\,eV\)
ⓓ. \(8.4\,eV\)
Correct Answer: \(1.6\,eV\)
Explanation: \( \textbf{Initial state:} \) The atom is already in \(n=2\).
\( \textbf{Ionisation energy from } n=2\textbf{:} \) \(3.4\,eV\).
\( \textbf{Photon energy supplied:} \) \(5.0\,eV\).
\( \textbf{Energy use:} \) First, \(3.4\,eV\) is used to free the electron from the atom.
\( \textbf{Remaining energy:} \)
\[
K=5.0\,eV-3.4\,eV
\]
\( \textbf{Calculation:} \)
\[
K=1.6\,eV
\]
\( \textbf{Physical meaning:} \) The electron is ionised, and the excess energy appears as kinetic energy.
\( \textbf{Final answer:} \) The kinetic energy of the freed electron is \(1.6\,eV\).
304. Use the energy-level description below.
A hydrogen atom has possible downward transitions from \(n=5\) to \(n=4\), from \(n=5\) to \(n=2\), and from \(n=5\) to \(n=1\).
The transition that emits the longest-wavelength photon is
ⓐ. \(5\to4\)
ⓑ. \(5\to2\)
ⓒ. \(5\to1\)
ⓓ. all three emit the same wavelength
Correct Answer: \(5\to4\)
Explanation: The longest wavelength corresponds to the smallest photon energy because \(E=\frac{hc}{\lambda}\). For downward transitions from the same initial level \(n=5\), the smallest energy drop occurs when the final level is closest to \(5\). The transition \(5\to4\) has a smaller energy difference than \(5\to2\) or \(5\to1\). Therefore, it emits the lowest-energy photon among the three. Lower photon energy means lower frequency and longer wavelength.
305. The hydrogen \(3\to2\) line has wavelength approximately \(656\,\text{nm}\). For the corresponding \(3\to2\) transition in \(He^+\), the wavelength is approximately
ⓐ. \(328\,\text{nm}\)
ⓑ. \(656\,\text{nm}\)
ⓒ. \(2624\,\text{nm}\)
ⓓ. \(164\,\text{nm}\)
Correct Answer: \(164\,\text{nm}\)
Explanation: \( \textbf{Hydrogen-like scaling:} \) For the same transition pair, photon energy is proportional to \(Z^2\).
\( \textbf{Hydrogen:} \) \(Z=1\).
\( \textbf{\(He^+\):} \) \(Z=2\).
\( \textbf{Energy ratio:} \)
\[
\frac{E_{He^+}}{E_H}=\frac{2^2}{1^2}=4
\]
\( \textbf{Wavelength relation:} \) Since \(E=\frac{hc}{\lambda}\), wavelength is inversely proportional to energy.
\( \textbf{Wavelength ratio:} \)
\[
\lambda_{He^+}=\frac{\lambda_H}{4}
\]
\( \textbf{Substitution:} \)
\[
\lambda_{He^+}=\frac{656\,\text{nm}}{4}=164\,\text{nm}
\]
\( \textbf{Final answer:} \) The corresponding \(He^+\) wavelength is approximately \(164\,\text{nm}\).
306. Compare the hydrogen-like species in the table.
| Species | \(Z\) | Same transition |
|---|
| P | \(1\) | \(3\to2\) |
| Q | \(2\) | \(3\to2\) |
| R | \(3\) | \(3\to2\) |
For the emitted photon energies, the correct order is
ⓐ. \(E_P\gt E_Q\gt E_R\)
ⓑ. \(E_P=E_Q=E_R\)
ⓒ. \(E_R\gt E_Q\gt E_P\)
ⓓ. \(E_Q\gt E_R\gt E_P\)
Correct Answer: \(E_R\gt E_Q\gt E_P\)
Explanation: For hydrogen-like species, the energy of a fixed transition is proportional to \(Z^2\). All three cases have the same transition \(3\to2\), so the bracket involving \(n_i\) and \(n_f\) is the same. The only changing factor is \(Z^2\). Case R has \(Z=3\), case Q has \(Z=2\), and case P has \(Z=1\). Therefore, the photon energy is largest for R, then Q, then P. The same transition label does not imply the same photon energy when the nuclear charge is different.
307. A table lists three Bohr-model quantities for hydrogen-like atoms.
| Row | Quantity | Dependence for fixed \(n\) |
|---|
| P | Orbit radius \(r_n\) | \(\propto \frac{1}{Z}\) |
| Q | Electron speed \(v_n\) | \(\propto Z\) |
| R | Magnitude of total energy \(|E_n|\) | \(\propto Z^2\) |
| S | Transition photon energy for same \(n_i\to n_f\) | \(\propto \frac{1}{Z^2}\) |
The row that needs correction is
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: For fixed \(n\), the radius formula \(r_n=\frac{n^2}{Z}a_0\) gives \(r_n\propto\frac{1}{Z}\). The speed formula \(v_n=\frac{Z}{n}v_{\mathrm{H},1}\) gives \(v_n\propto Z\). The energy formula \(E_n=-13.6\frac{Z^2}{n^2}\,eV\) shows that the magnitude \(|E_n|\) is proportional to \(Z^2\). For a fixed transition \(n_i\to n_f\), the energy difference also carries the same \(Z^2\) factor. Row S reverses this dependence, so it conflicts with hydrogen-like ion scaling.
308. A photon of wavelength \(486\,\text{nm}\) corresponds to a hydrogen Balmer transition. Using \(hc=1240\,eV\,\text{nm}\), its energy is closest to
ⓐ. \(2.55\,eV\)
ⓑ. \(1.89\,eV\)
ⓒ. \(3.40\,eV\)
ⓓ. \(10.20\,eV\)
Correct Answer: \(2.55\,eV\)
Explanation: \( \textbf{Given wavelength:} \) \(\lambda=486\,\text{nm}\).
\( \textbf{Photon-energy relation:} \)
\[
E=\frac{hc}{\lambda}
\]
\( \textbf{Given constant form:} \)
\[
hc=1240\,eV\,\text{nm}
\]
\( \textbf{Substitution:} \)
\[
E=\frac{1240}{486}\,eV
\]
\( \textbf{Calculation:} \)
\[
E\approx2.55\,eV
\]
\( \textbf{Hydrogen-level link:} \) This matches the energy difference for the \(4\to2\) Balmer transition.
\( \textbf{Final answer:} \) The photon energy is closest to \(2.55\,eV\).
309. A hydrogen atom emits a photon of frequency \(4.57\times10^{14}\,\text{Hz}\). Taking \(h=6.6\times10^{-34}\,\text{J s}\) and \(1\,eV=1.6\times10^{-19}\,\text{J}\), the photon energy is closest to
ⓐ. \(1.89\,eV\)
ⓑ. \(0.94\,eV\)
ⓒ. \(3.40\,eV\)
ⓓ. \(10.2\,eV\)
Correct Answer: \(1.89\,eV\)
Explanation: \( \textbf{Given frequency:} \) \(\nu=4.57\times10^{14}\,\text{Hz}\).
\( \textbf{Planck relation:} \)
\[
E=h\nu
\]
\( \textbf{Substitution in joule:} \)
\[
E=(6.6\times10^{-34})(4.57\times10^{14})\,\text{J}
\]
\( \textbf{Power of ten:} \)
\[
10^{-34}\times10^{14}=10^{-20}
\]
\( \textbf{Numerical product:} \)
\[
6.6\times4.57\approx30.16
\]
\( \textbf{Energy in joule:} \)
\[
E\approx30.16\times10^{-20}\,\text{J}=3.016\times10^{-19}\,\text{J}
\]
\( \textbf{Conversion to } eV\textbf{:} \)
\[
E=\frac{3.016\times10^{-19}}{1.6\times10^{-19}}\,eV\approx1.89\,eV
\]
\( \textbf{Final answer:} \) The photon energy is approximately \(1.89\,eV\).
310. A hydrogen atom is excited to \(n=5\). The maximum number of distinct emission lines possible as atoms in a sample return to lower levels is
ⓐ. \(5\)
ⓑ. \(10\)
ⓒ. \(8\)
ⓓ. \(20\)
Correct Answer: \(10\)
Explanation: \( \textbf{Highest excited level:} \) \(n=5\).
\( \textbf{Line-count formula:} \)
\[
N=\frac{n(n-1)}{2}
\]
\( \textbf{Substitution:} \)
\[
N=\frac{5(5-1)}{2}
\]
\( \textbf{Simplification:} \)
\[
N=\frac{5\times4}{2}=10
\]
\( \textbf{Meaning:} \) The count includes all possible downward transition pairs among levels \(5,4,3,2,\) and \(1\).
\( \textbf{Sample interpretation:} \) A collection of atoms can show all these lines through different cascade paths.
\( \textbf{Final answer:} \) The maximum number of distinct emission lines is \(10\).
311. A single hydrogen atom at \(n=5\) falls directly to \(n=2\). The emitted photon belongs to the
ⓐ. Lyman series
ⓑ. Paschen series
ⓒ. Brackett series
ⓓ. Balmer series
Correct Answer: Balmer series
Explanation: A hydrogen spectral series is identified by the final lower level. The transition \(5\to2\) ends at \(n=2\). All transitions ending at \(n=2\) are members of the Balmer series. The starting level \(n=5\) identifies which Balmer line is produced, but it does not change the series name. Confusing the initial level with the final level gives wrong series identification.
312. Use the graph description below.
A graph of \(\frac{1}{\lambda}\) versus \(\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) is drawn for hydrogen spectral lines. The plotted points lie on a straight line through the origin.
The slope of this graph represents
ⓐ. Bohr radius \(a_0\)
ⓑ. Planck constant \(h\)
ⓒ. Rydberg constant \(R\)
ⓓ. electron speed \(v_n\)
Correct Answer: Rydberg constant \(R\)
Explanation: The Rydberg formula for hydrogen is
\[
\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\]
This has the straight-line form \(y=mx\). Here \(y=\frac{1}{\lambda}\), \(x=\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\), and the slope \(m\) is \(R\). The bracket is dimensionless, while \(\frac{1}{\lambda}\) has unit \(\text{m}^{-1}\), so the slope must also have unit \(\text{m}^{-1}\). The graph connects spectral data with the Rydberg constant, not with orbital radius directly.
313. For a hydrogen-like ion, the graph of \(\frac{1}{\lambda}\) versus \(\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) has slope \(RZ^2\). If the slope is \(9R\), the ion has
ⓐ. \(Z=1\)
ⓑ. \(Z=2\)
ⓒ. \(Z=3\)
ⓓ. \(Z=9\)
Correct Answer: \(Z=3\)
Explanation: \( \textbf{Hydrogen-like Rydberg form:} \)
\[
\frac{1}{\lambda}=RZ^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
\]
\( \textbf{Graph slope:} \) The slope is \(RZ^2\).
\( \textbf{Given slope:} \)
\[
RZ^2=9R
\]
\( \textbf{Cancel } R\textbf{:} \)
\[
Z^2=9
\]
\( \textbf{Solve for positive } Z\textbf{:} \)
\[
Z=3
\]
\( \textbf{Physical meaning:} \) This corresponds to a hydrogen-like ion with nuclear charge \(+3e\), such as \(Li^{2+}\).
\( \textbf{Final answer:} \) The ion has \(Z=3\).
314. The de Broglie wavelength of an electron moving with momentum \(p\) is
ⓐ. \(\lambda=\frac{h}{p}\)
ⓑ. \(\lambda=hp\)
ⓒ. \(\lambda=\frac{p}{h}\)
ⓓ. \(\lambda=\frac{h}{p^2}\)
Correct Answer: \(\lambda=\frac{h}{p}\)
Explanation: De Broglie associated a wavelength with a moving particle. The relation is \(\lambda=\frac{h}{p}\), where \(p\) is the particle’s momentum. For an electron of mass \(m_e\) moving with speed \(v\), this may be written as \(\lambda=\frac{h}{m_ev}\) when non-relativistic momentum is used. This matter-wave idea helps explain why only certain Bohr orbits are allowed. The wavelength is inversely proportional to momentum, so larger momentum means shorter matter wavelength.
315. In the de Broglie interpretation of Bohr orbits, an allowed circular orbit is one in which
ⓐ. half a wavelength is always forbidden from every orbit
ⓑ. the electron wavelength is unrelated to the orbit radius
ⓒ. integer electron wavelengths fit around the orbit
ⓓ. the electron emits radiation continuously along the orbit
Correct Answer: integer electron wavelengths fit around the orbit
Explanation: In the wave interpretation, the electron behaves like a matter wave around the circular orbit. A stable standing wave is formed only if the circumference contains a whole number of wavelengths. This condition is written as \(2\pi r_n=n\lambda\), where \(n=1,2,3,\ldots\). If the wave does not fit smoothly around the circle, it would not reproduce itself consistently after one round. The standing-wave condition gives a physical meaning to the allowed orbit number \(n\).
316. A circular electron orbit has circumference \(2\pi r=5\lambda\). In the de Broglie standing-wave picture, the orbit corresponds to
ⓐ. \(n=1\)
ⓑ. \(n=2\)
ⓒ. \(n=10\)
ⓓ. \(n=5\)
Correct Answer: \(n=5\)
Explanation: The standing-wave condition for a Bohr orbit is
\[
2\pi r=n\lambda
\]
\( \textbf{Given condition:} \)
\[
2\pi r=5\lambda
\]
\( \textbf{Comparison with standard form:} \)
\[
n\lambda=5\lambda
\]
\( \textbf{Allowed orbit number:} \)
\[
n=5
\]
\( \textbf{Meaning:} \) Five complete electron matter wavelengths fit around the circumference.
\( \textbf{Final answer:} \) The orbit corresponds to \(n=5\).
317. If an electron wave around a proposed circular orbit contains \(3.5\) wavelengths around the circumference, the orbit is not allowed in the Bohr-de Broglie picture because
ⓐ. the electron has no momentum in that orbit
ⓑ. the nucleus becomes electrically neutral
ⓒ. the wave does not close smoothly on itself
ⓓ. the angular momentum becomes exactly \(3\hbar\)
Correct Answer: the wave does not close smoothly on itself
Explanation: An allowed orbit requires the electron matter wave to form a standing wave around the circle. This needs \(2\pi r=n\lambda\), where \(n\) is a positive integer. A value such as \(3.5\) wavelengths means the wave would return out of phase after one complete round. Such a wave would not join smoothly with itself around the circumference. The non-integer fit is the reason the orbit is excluded, not a loss of electron momentum.
318. Combining \(2\pi r=n\lambda\) with \(\lambda=\frac{h}{m_ev}\) gives Bohr’s angular momentum condition. The correct intermediate substitution is
ⓐ. \(2\pi r=n\frac{m_ev}{h}\)
ⓑ. \(2\pi r=\frac{h}{nm_ev}\)
ⓒ. \(2\pi r=\frac{m_evh}{n}\)
ⓓ. \(2\pi r=n\frac{h}{m_ev}\)
Correct Answer: \(2\pi r=n\frac{h}{m_ev}\)
Explanation: \( \textbf{Standing-wave condition:} \)
\[
2\pi r=n\lambda
\]
\( \textbf{de Broglie relation:} \)
\[
\lambda=\frac{h}{m_ev}
\]
\( \textbf{Substitution step:} \)
\[
2\pi r=n\frac{h}{m_ev}
\]
\( \textbf{Next step:} \) Multiplying both sides by \(m_ev\) gives \(2\pi m_evr=nh\).
\( \textbf{Final rearrangement:} \)
\[
m_evr=\frac{nh}{2\pi}
\]
\( \textbf{Final answer:} \) The correct substitution is \(2\pi r=n\frac{h}{m_ev}\).
319. Assertion: The de Broglie standing-wave condition gives a physical interpretation of Bohr’s angular momentum quantisation.
Reason: \(2\pi r=n\lambda\) together with \(\lambda=\frac{h}{mv}\) leads to \(mvr=\frac{nh}{2\pi}\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Bohr originally introduced angular momentum quantisation as a postulate. The de Broglie idea later connected that postulate with wave behaviour of the electron. If an integral number of wavelengths fits around the orbit, \(2\pi r=n\lambda\). Substituting \(\lambda=\frac{h}{mv}\) gives \(mvr=\frac{nh}{2\pi}\). The Reason states the mathematical bridge from standing matter waves to the quantised angular momentum condition.
320. Study the table about de Broglie and Bohr quantisation.
| Row | Statement | Status |
|---|
| P | \(\lambda=\frac{h}{p}\) | de Broglie wavelength relation |
| Q | \(2\pi r=n\lambda\) | standing-wave condition for allowed circular orbit |
| R | \(mvr=\frac{nh}{2\pi}\) | Bohr angular momentum condition |
| S | \(2\pi r=3.5\lambda\) | allowed standing wave with \(n=3.5\) |
The row that needs correction is
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: The de Broglie wavelength relation is \(\lambda=\frac{h}{p}\), so row P is correct. The condition \(2\pi r=n\lambda\) gives a standing wave around a circular orbit, so row Q is correct. Combining that condition with \(\lambda=\frac{h}{mv}\) gives \(mvr=\frac{nh}{2\pi}\), so row R is also correct. Row S is wrong because the number of wavelengths around an allowed orbit must be a positive integer. A non-integer value such as \(3.5\) does not produce a smoothly closing standing wave.
