301. Bohr introduced the idea of stationary states mainly to explain why an electron in an allowed orbit
ⓐ. continuously radiates energy while revolving
ⓑ. does not radiate in a permitted orbit
ⓒ. changes into a neutron inside the nucleus
ⓓ. has no interaction with the nucleus
Correct Answer: does not radiate in a permitted orbit
Explanation: Rutherford's model had a serious stability problem because a revolving charged electron should radiate energy according to classical theory. Bohr avoided this by postulating that electrons can revolve only in certain allowed or stationary orbits. While an electron remains in one of these stationary states, it does not emit radiation. Radiation is emitted or absorbed only when the electron jumps between allowed energy levels. This postulate was introduced to account for atomic stability and line spectra, not to remove the electron-nucleus attraction.
302. The angular momentum condition in Bohr's model is written as
ⓐ. \(mvr=\frac{nh}{2\pi}\), where \(n=1,2,3,\ldots\)
ⓑ. \(mvr=\frac{2\pi}{nh}\), where \(n=0,\frac{1}{2},1,\ldots\)
ⓒ. \(mvr=nh^2\), where \(n\) may be any decimal
ⓓ. \(mvr=\frac{h}{n^2}\), where \(n\) is always \(0\)
Correct Answer: \(mvr=\frac{nh}{2\pi}\), where \(n=1,2,3,\ldots\)
Explanation: Bohr proposed that the angular momentum of an electron in an allowed orbit is quantized. The allowed values are integral multiples of \(\frac{h}{2\pi}\). This is written as \(mvr=\frac{nh}{2\pi}\), where \(n\) is a positive integer. The value \(n=0\) is not used for an orbit in Bohr's model because it would give zero angular momentum and no physical orbit. The integer condition is the quantization part of the postulate.
303. In Bohr's model, radiation is emitted by an atom when an electron
ⓐ. jumps from a higher energy level to a lower energy level
ⓑ. stays permanently in the same stationary orbit
ⓒ. jumps from a lower energy level to a higher energy level
ⓓ. moves in any circular path with no energy change
Correct Answer: jumps from a higher energy level to a lower energy level
Explanation: In Bohr's model, an electron does not radiate energy while remaining in a stationary orbit. Radiation is connected with transitions between allowed energy levels. When an electron falls from a higher energy level to a lower energy level, the energy difference is emitted as a photon. The photon energy is given by \(\Delta E=h\nu\). A jump from lower to higher level requires absorption of energy, so the transition direction decides whether energy is emitted or absorbed.
304. Two statements about Bohr's postulates are given.
Statement I: Electrons in allowed stationary orbits do not radiate energy.
Statement II: The angular momentum of an electron in an allowed orbit can have any arbitrary value.
ⓐ. Both Statement I and Statement II are true
ⓑ. Statement I is true, but Statement II is false
ⓒ. Statement I is false, but Statement II is true
ⓓ. Both Statement I and Statement II are false
Correct Answer: Statement I is true, but Statement II is false
Explanation: Bohr's stationary-state postulate says that an electron does not radiate energy while it remains in an allowed orbit. This was used to explain the stability of atoms. However, Bohr did not allow angular momentum to take any arbitrary value. He restricted it to \(mvr=\frac{nh}{2\pi}\), where \(n\) is a positive integer. Statement II misses the quantization condition, which is central to Bohr's model.
305. A hydrogen electron is proposed to have angular momentum equal to \(\frac{5h}{2\pi}\). According to Bohr's quantization rule, the corresponding orbit has
ⓐ. \(n=2\)
ⓑ. \(n=\frac{1}{5}\)
ⓒ. \(n=5\)
ⓓ. \(n=0\)
Correct Answer: \(n=5\)
Explanation: \( \textbf{Given angular momentum:} \) \(mvr=\frac{5h}{2\pi}\).
\( \textbf{Bohr quantization rule:} \)
\[
mvr=\frac{nh}{2\pi}
\]
\( \textbf{Compare the two forms:} \)
\[
\frac{nh}{2\pi}=\frac{5h}{2\pi}
\]
\( \textbf{Cancel the common factor:} \) The factor \(\frac{h}{2\pi}\) is common on both sides.
\( \textbf{Value of }n\textbf{:} \)
\[
n=5
\]
\( \textbf{Allowed-orbit check:} \) \(n=5\) is a positive integer, so it is allowed in Bohr's model.
\( \textbf{Final answer:} \) The orbit corresponds to \(n=5\). The coefficient multiplying \(\frac{h}{2\pi}\) directly gives the quantum number \(n\).
306. Read the situation below and answer the question.
An electron in a hydrogen atom first stays in level \(n=2\). It then absorbs a photon and moves to level \(n=4\).
This change is best described as
ⓐ. emission because the electron moves away from the nucleus
ⓑ. no energy exchange because both levels are stationary states
ⓒ. nuclear reaction because the value of \(n\) changes
ⓓ. absorption; the electron moves to a higher level
Correct Answer: absorption; the electron moves to a higher level
Explanation: Levels \(n=2\) and \(n=4\) are stationary states, but a jump between them still involves energy exchange. Moving from \(n=2\) to \(n=4\) means the electron goes to a higher energy level. Such a transition requires absorption of a photon with energy equal to the difference between the two levels. Emission would occur for the reverse transition, such as \(n=4\rightarrow n=2\). The change is an electronic transition, not a nuclear change.
307. Assertion: Bohr's model explains atomic stability better than Rutherford's model.
Reason: Bohr postulated that electrons in stationary orbits do not radiate energy.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. 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
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Rutherford's model could not explain why revolving electrons did not continuously lose energy and collapse into the nucleus. Bohr introduced stationary orbits in which electrons do not radiate energy. This postulate directly removed the classical collapse problem for allowed orbits. The Assertion is true because Bohr's model improved the stability explanation. The Reason is also true and gives the specific postulate responsible for that improvement.
308. The row that correctly connects a Bohr-model idea with its meaning is
| Row | Bohr-model idea | Meaning |
|---|
| P | Stationary orbit | Electron radiates continuously in it |
| Q | Angular momentum quantization | \(mvr=\frac{nh}{2\pi}\) |
| R | Emission transition | Electron jumps from lower \(n\) to higher \(n\) |
| S | Absorption transition | Electron falls from higher \(n\) to lower \(n\) |
ⓐ. P
ⓑ. R
ⓒ. Q
ⓓ. S
Correct Answer: Q
Explanation: Angular momentum quantization in Bohr's model is represented by \(mvr=\frac{nh}{2\pi}\). Row \(P\) is wrong because an electron does not radiate while remaining in a stationary orbit. Row \(R\) is wrong because emission occurs when the electron moves from a higher level to a lower level. Row \(S\) is wrong because absorption occurs when the electron moves from lower level to higher level. The table shows why transition direction and stationary-state behaviour must be kept separate.
309. If an electron emits a photon of frequency \(\nu\) during a Bohr transition, the energy difference between the two levels is
ⓐ. \(\Delta E=\frac{\nu}{h}\)
ⓑ. \(\Delta E=h\nu\)
ⓒ. \(\Delta E=h+\nu\)
ⓓ. \(\Delta E=\frac{h}{\nu^2}\)
Correct Answer: \(\Delta E=h\nu\)
Explanation: Bohr connected spectral lines with electron jumps between allowed energy levels. The energy of the emitted or absorbed photon equals the energy difference between the two levels. Planck's relation gives photon energy as \(E=h\nu\). Therefore, the level difference is written as \(\Delta E=h\nu\). The relation is linear in frequency, so doubling frequency doubles the energy difference.
310. For hydrogen, Bohr's radius relation is \(r_n=a_0n^2\). The radius of the third orbit is
ⓐ. \(3a_0\)
ⓑ. \(6a_0\)
ⓒ. \(\frac{a_0}{9}\)
ⓓ. \(9a_0\)
Correct Answer: \(9a_0\)
Explanation: \( \textbf{Given relation:} \) \(r_n=a_0n^2\).
\( \textbf{Required orbit:} \) Third orbit means \(n=3\).
\( \textbf{Substitution:} \)
\[
r_3=a_0(3)^2
\]
\( \textbf{Square step:} \)
\[
3^2=9
\]
\( \textbf{Radius:} \)
\[
r_3=9a_0
\]
\( \textbf{Final answer:} \) The third Bohr orbit has radius \(9a_0\). The radius is proportional to \(n^2\), not directly to \(n\).
311. The radius of the second Bohr orbit of hydrogen is closest to \(\left(a_0=52.9\,\text{pm}\right)\)
ⓐ. \(26.45\,\text{pm}\)
ⓑ. \(211.6\,\text{pm}\)
ⓒ. \(52.9\,\text{pm}\)
ⓓ. \(105.8\,\text{pm}\)
Correct Answer: \(211.6\,\text{pm}\)
Explanation: \( \textbf{Given Bohr radius:} \) \(a_0=52.9\,\text{pm}\).
\( \textbf{Hydrogen radius relation:} \)
\[
r_n=a_0n^2
\]
\( \textbf{For the second orbit:} \) \(n=2\).
\( \textbf{Substitution:} \)
\[
r_2=52.9(2)^2\,\text{pm}
\]
\( \textbf{Square of }2\textbf{:} \)
\[
2^2=4
\]
\( \textbf{Calculation:} \)
\[
r_2=52.9\times4=211.6\,\text{pm}
\]
\( \textbf{Final answer:} \) The radius is \(211.6\,\text{pm}\). Multiplying by \(2\) instead of \(2^2\) would miss the square dependence.
312. Use the graph description below.
For hydrogen, a graph is plotted with Bohr orbit radius \(r_n\) on the y-axis and \(n^2\) on the x-axis.
The graph should be
ⓐ. a decreasing curve with negative slope
ⓑ. a horizontal line because radius is independent of \(n\)
ⓒ. a straight line with slope \(-13.6\,\text{eV}\)
ⓓ. a straight line through the origin with slope \(a_0\)
Correct Answer: a straight line through the origin with slope \(a_0\)
Explanation: For hydrogen, the radius relation is \(r_n=a_0n^2\). If \(r_n\) is plotted against \(n^2\), this relation has the straight-line form \(y=mx\). The slope of the line is \(a_0\), and the line passes through the origin in the mathematical relation. The value \(-13.6\,\text{eV}\) belongs to the energy relation, not the radius relation. The graph tests the square dependence of orbit size on \(n\).
313. Two hydrogen orbits have principal quantum numbers \(n=2\) and \(n=5\). The ratio \(\frac{r_5}{r_2}\) is
ⓐ. \(\frac{25}{4}\)
ⓑ. \(\frac{5}{2}\)
ⓒ. \(\frac{4}{25}\)
ⓓ. \(\frac{2}{5}\)
Correct Answer: \(\frac{25}{4}\)
Explanation: \( \textbf{Radius relation for hydrogen:} \)
\[
r_n=a_0n^2
\]
\( \textbf{Write the ratio:} \)
\[
\frac{r_5}{r_2}=\frac{a_0(5)^2}{a_0(2)^2}
\]
\( \textbf{Cancel }a_0\textbf{:} \)
\[
\frac{r_5}{r_2}=\frac{25}{4}
\]
\( \textbf{Meaning:} \) The fifth orbit is \(\frac{25}{4}\) times as large in radius as the second orbit.
\( \textbf{Final answer:} \) \(\frac{r_5}{r_2}=\frac{25}{4}\). The ratio depends on the squares of quantum numbers, not on the quantum numbers directly.
314. A student claims, "In hydrogen, the third Bohr orbit is three times as far from the nucleus as the first orbit." The correction is that the third orbit is
ⓐ. \(3\) times as far because \(r_n\propto n\)
ⓑ. \(6\) times as far because \(r_n\propto2n\)
ⓒ. \(9\) times as far because \(r_n\propto n^2\)
ⓓ. \(\frac{1}{3}\) as far because radius decreases with \(n\)
Correct Answer: \(9\) times as far because \(r_n\propto n^2\)
Explanation: The hydrogen radius relation is \(r_n=a_0n^2\). For \(n=1\), the radius is \(r_1=a_0\). For \(n=3\), the radius is \(r_3=9a_0\). Therefore, \(\frac{r_3}{r_1}=9\). The student's statement treats radius as directly proportional to \(n\), but Bohr's relation gives a square dependence.
315. A hydrogen-like species has nuclear charge \(Z\). Its Bohr radius relation is \(r_n=\frac{a_0n^2}{Z}\). For the same \(n\), increasing \(Z\) causes the orbit radius to
ⓐ. increase in direct proportion to \(Z\) in the notation
ⓑ. remain unchanged for all one-electron species
ⓒ. become negative in the notation
ⓓ. decrease as nuclear attraction becomes stronger
Correct Answer: decrease as nuclear attraction becomes stronger
Explanation: For hydrogen-like species, the orbit radius is \(r_n=\frac{a_0n^2}{Z}\). At fixed \(n\), the radius is inversely proportional to \(Z\). A higher nuclear charge pulls the one electron more strongly, so the allowed orbit is smaller. This is why \(\mathrm{He^+}\) has a smaller radius than \(\mathrm{H}\) for the same principal quantum number. The radius remains positive; only the size changes with \(Z\).
316. The energy of the \(n\)th level of hydrogen in Bohr's model is \(E_n=-\frac{13.6}{n^2}\,\text{eV}\). The negative sign means that the electron
ⓐ. is bound relative to the separated electron state
ⓑ. has no energy at all in the Bohr model for the electron
ⓒ. has negative mass in the Bohr model for the electron
ⓓ. must always emit radiation in the same orbit
Correct Answer: is bound relative to the separated electron state
Explanation: In Bohr's model, the energy of a bound electron in hydrogen is taken as negative relative to the zero energy of a completely separated electron at \(n=\infty\). The negative sign does not mean the electron has negative mass or no energy. It means energy must be supplied to remove the electron from the atom. As \(n\) increases, \(E_n\) becomes less negative and approaches \(0\). The sign convention is about binding, not about the charge of the electron.
317. For hydrogen, the energy of the level \(n=2\) is
ⓐ. \(-13.6\,\text{eV}\)
ⓑ. \(-6.8\,\text{eV}\)
ⓒ. \(-3.4\,\text{eV}\)
ⓓ. \(+3.4\,\text{eV}\)
Correct Answer: \(-3.4\,\text{eV}\)
Explanation: \( \textbf{Energy relation for hydrogen:} \)
\[
E_n=-\frac{13.6}{n^2}\,\text{eV}
\]
\( \textbf{For }n=2\textbf{:} \)
\[
E_2=-\frac{13.6}{2^2}\,\text{eV}
\]
\( \textbf{Square step:} \)
\[
2^2=4
\]
\( \textbf{Calculation:} \)
\[
E_2=-\frac{13.6}{4}=-3.4\,\text{eV}
\]
\( \textbf{Final answer:} \) \(E_2=-3.4\,\text{eV}\). The energy remains negative because the electron is still bound in the atom.
318. The energy of a hydrogen electron changes from \(E_3=-1.51\,\text{eV}\) to \(E_2=-3.40\,\text{eV}\). The transition is
ⓐ. absorption of \(1.89\,\text{eV}\)
ⓑ. emission of \(-4.91\,\text{eV}\)
ⓒ. absorption of \(4.91\,\text{eV}\)
ⓓ. emission of \(1.89\,\text{eV}\)
Correct Answer: emission of \(1.89\,\text{eV}\)
Explanation: \( \textbf{Initial level:} \) \(E_3=-1.51\,\text{eV}\).
\( \textbf{Final level:} \) \(E_2=-3.40\,\text{eV}\).
\( \textbf{Transition direction:} \) The electron moves from \(n=3\) to \(n=2\), a lower energy level.
\( \textbf{Energy released:} \)
\[
E_{\text{photon}}=E_{\text{initial}}-E_{\text{final}}
\]
\( \textbf{Substitution:} \)
\[
E_{\text{photon}}=(-1.51)-(-3.40)\,\text{eV}
\]
\( \textbf{Calculation:} \)
\[
E_{\text{photon}}=1.89\,\text{eV}
\]
\( \textbf{Final answer:} \) The atom emits a photon of energy \(1.89\,\text{eV}\). Although the level energies are negative, the emitted photon energy is reported as a positive energy.
319. A hydrogen atom in the ground state has \(E_1=-13.6\,\text{eV}\). The energy required to ionize it from the ground state is
ⓐ. \(0\,\text{eV}\)
ⓑ. \(3.4\,\text{eV}\)
ⓒ. \(13.6\,\text{eV}\)
ⓓ. \(10.2\,\text{eV}\)
Correct Answer: \(13.6\,\text{eV}\)
Explanation: Ionization from the ground state means removing the electron from \(n=1\) to \(n=\infty\). In Bohr's energy scale, the energy at \(n=\infty\) is \(0\,\text{eV}\). The ground-state energy is \(-13.6\,\text{eV}\). The required energy is \(0-(-13.6)=13.6\,\text{eV}\). The energy supplied must exactly overcome the binding energy of the electron in the ground state.
320. A graph is drawn with \(E_n\) of hydrogen on the y-axis and \(\frac{1}{n^2}\) on the x-axis. Since \(E_n=-13.6\left(\frac{1}{n^2}\right)\,\text{eV}\), the graph should be
ⓐ. a straight line with positive slope \(+13.6\,\text{eV}\)
ⓑ. a straight line with negative slope
ⓒ. a horizontal line at \(+13.6\,\text{eV}\)
ⓓ. a curve showing \(E_n\) increasing without limit above zero
Correct Answer: a straight line with negative slope
Explanation: The hydrogen energy relation can be written as \(E_n=-13.6\left(\frac{1}{n^2}\right)\,\text{eV}\). This has the form \(y=mx\), where the slope is \(-13.6\,\text{eV}\). Therefore, a plot of \(E_n\) against \(\frac{1}{n^2}\) is a straight line with negative slope. As \(n\) increases, \(\frac{1}{n^2}\) decreases and \(E_n\) approaches \(0\) from the negative side. The graph should not cross into positive energy for bound hydrogen levels.
