Q.State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum does it belong.

Showing the 12 most recent of 99 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.For questions 13 to 16, two statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false. Assertion (A) : In Bohr model of hydrogen atom, the energy levels are discrete and quantised. Reason (R) : In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.

   ›Reveal solutionSolution

   The key idea is that Bohr’s model indeed has discrete energy levels (true), and the reason given — that electrostatic force provides centripetal force — is also true, but it is a classical condition that does not explain quantisation. So both are true, but the reason is not the correct explanation of the assertion. The answer is (B).

   Let’s unpack this carefully. The question is from the Assertion-Reason format, so we need to check two things independently: whether each statement is true, and then whether the reason correctly explains the assertion.

   1. Check the Assertion (A):

   “In Bohr model of hydrogen atom, the energy levels are discrete and quantised.”

   This is absolutely true. Bohr’s model was revolutionary precisely because it introduced quantised orbits — electrons can only occupy certain allowed energy levels, and they jump between these levels by absorbing or emitting photons of specific energies. The energy of the $n$-th level is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$, which is clearly discrete. So Assertion (A) is true.

   2. Check the Reason (R):

   “In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.”

   This is also true — it’s a standard Newtonian condition for any circular orbit. For an electron of mass $m$ and speed $v$ at a distance $r$ from the nucleus, the Coulomb attraction $\frac{ke^2}{r^2}$ equals the centripetal force $m v^2 / r$. So Reason (R) is true.

   3. Now the crucial part: Does (R) correctly explain (A)?

   The reason describes a classical force balance — it’s the same condition that would hold for a planet orbiting the Sun. But that classical condition alone does not lead to discrete, quantised energy levels. In fact, classical physics would allow any orbit radius, and hence any energy. Bohr had to add an extra postulate — quantisation of angular momentum ($m v r = n \hbar$) — to get discrete levels. The electrostatic-centripetal balance is necessary for the model, but it is not the reason for quantisation. So (R) is true but does not explain (A). …

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2. CBSE 2026Set 55/2/11 markMCQ
   Q.In Bohr model of hydrogen atom, for large values of $n$, the distance between the consecutive orbits is proportional to (A) $\sqrt{n}$ (B) $n$ (C) $n^2$ (D) $n^3$

   ›Reveal solutionSolution

   In the Bohr model, the orbital radius scales as $r_n \propto n^2$. For large $n$, the gap between consecutive orbits $\Delta r = r_{n+1} - r_n$ behaves like $2n + 1$, which is proportional to $n$ — so the answer is (B).

   The Bohr model gives a beautifully simple picture of the hydrogen atom: electrons orbit the nucleus in fixed circular paths, with quantised angular momentum. The radius of the $n$-th orbit is

   $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \quad \text{or, more compactly,} \quad r_n = a_0 n^2,$

   where $a_0 \approx 0.529 \, \text{Å}$ is the Bohr radius. So the radius grows as $n^2$.

   The question asks about the distance between consecutive orbits — that is, the difference $r_{n+1} - r_n$ — for large $n$. Many students instinctively think this difference is constant, or that it grows like $n^2$ because the radii themselves do. But a difference between two quadratic terms behaves differently from either term alone. Let’s work it out.

   1. Write the radii for two neighbouring orbits:

   $r_n = a_0 n^2, \qquad r_{n+1} = a_0 (n+1)^2.$

   2. The gap between them is

   $\Delta r = r_{n+1} - r_n = a_0 \left[ (n+1)^2 - n^2 \right].$

   3. Expand $(n+1)^2 = n^2 + 2n + 1$. Then

   $\Delta r = a_0 \left( n^2 + 2n + 1 - n^2 \right) = a_0 (2n + 1).$

   4. For large $n$, the constant $1$ becomes negligible compared to $2n$. So

   $\Delta r \approx 2 a_0 n.$

   Thus the spacing between consecutive orbits is proportional to $n$ itself — not $n^2$, not $\sqrt{n}$, not $n^3$. …

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3. CBSE 2026Set 55/2/11 markMCQ
   Q.In Bohr model of hydrogen atom, the electron makes a transition from $n = 5$ to $n = 1$ state. As a result, a photon of wavelength $\lambda$ is emitted. The wavelength of the photon emitted when an electron makes a transition from energy level $n = 5$ to $n = 2$ will be (A) $\dfrac{8}{7}\lambda$ (B) $\dfrac{24}{7}\lambda$ (C) $\dfrac{16}{7}\lambda$ (D) $\dfrac{32}{7}\lambda$

   ›Reveal solutionSolution

   The energy difference between levels determines photon wavelength through $E = \frac{hc}{\lambda}$. Since the $5 \to 2$ transition releases less energy than $5 \to 1$, its photon has a longer wavelength. The answer is $\boxed{\frac{32}{7}\lambda}$ — option (D).

   The Bohr model tells us that when an electron drops from a higher energy level to a lower one, it emits a photon whose energy exactly equals the energy difference between those levels. The key relationship is $E_{\text{photon}} = \frac{hc}{\lambda}$, which shows that energy and wavelength are inversely related: a smaller energy gap produces a longer wavelength.

   In hydrogen, the energy of the $n$-th level is given by:

   $E_n = -\frac{13.6 \text{ eV}}{n^2}$

   The negative sign indicates that the electron is bound to the nucleus. When the electron transitions from level $n_i$ to $n_f$, the energy released is:

   $\Delta E = E_{n_i} - E_{n_f} = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ eV}$

   This energy becomes the photon's energy: $\Delta E = \frac{hc}{\lambda}$.

   For the first transition ($5 \to 1$):

   1. Calculate the energy difference:

   $\Delta E_1 = 13.6 \left(\frac{1}{1^2} - \frac{1}{5^2}\right) = 13.6 \left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25}$

   2. This energy corresponds to wavelength $\lambda$:

   $\frac{hc}{\lambda} = 13.6 \times \frac{24}{25}$

   For the second transition ($5 \to 2$):

   3. Calculate the energy difference:

   $\Delta E_2 = 13.6 \left(\frac{1}{2^2} - \frac{1}{5^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{25}\right)$

   4. Find a common denominator:

   $\Delta E_2 = 13.6 \left(\frac{25 - 4}{100}\right) = 13.6 \times \frac{21}{100}$

   5. This energy corresponds to wavelength $\lambda'$: …

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4. CBSE 2026Set V11 markMCQ
   Q.Let K be the kinetic energy, U be the potential energy and E be the total energy of an electron revolving around the nucleus in a hydrogen atom, then which of the following is correct?

   (a) $K > 0,\ U > 0,\ E > 0$

   (b) $K > 0,\ U < 0,\ E < 0$

   (c) $K > 0,\ U > 0,\ E < 0$

   (d) $K < 0,\ U < 0,\ E < 0$

   ›Reveal solutionSolution

   Option (b) $K > 0,\ U < 0,\ E < 0$. …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Formula for total energy of the electron in the nth stationary state of the hydrogen atom is -

   (i) $-\dfrac{11.2}{n^2}$ eV

   (ii) $+\dfrac{11.2}{n^2}$ eV

   (iii) $-\dfrac{13.6}{n^2}$ eV

   (iv) $+\dfrac{13.6}{n^2}$ eV

   ›Reveal solutionSolution

   Total energy of the electron in the nth Bohr orbit of hydrogen is $-13.6/n^2$ eV.

   …

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6. CBSE 2026Set A1 markMCQ
   Q.The solar spectrum is (A) continuous (B) line spectrum (C) spectrum of black lines (D) spectrum of black bands

   ›Reveal solutionSolution

   The solar spectrum is a line-absorption spectrum: a continuous background crossed by dark (Fraunhofer) lines.

   The hot, dense interior of the Sun emits a continuous spectrum. As this light passes through the cooler gases of the Sun's outer atmosphere, atoms there absorb their characteristic wavelengths, leaving dark lines (Fraunhofer lines) superimposed on the …

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7. CBSE 2026Set A1 markMCQ
   Q.The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2

   ›Reveal solutionSolution

   Bohr radii scale as n², so r₂ = 4 r₀.

   In Bohr's model of hydrogen the radius of the $n$-th orbit is:

   $r_n=n^2 r_0$

   …

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8. CBSE 2026Set A1 markMCQ
   Q.The kinetic energy (K) of an electron in a Bohr orbit is related to its potential energy (U) by (A) K = U (B) K = -U (C) K = -U/2 (D) K = -2U

   ›Reveal solutionSolution

   For an electron in a Bohr orbit, K = -U/2 (with U taken negative), which also gives total energy E = -K.

   In a hydrogen-like Bohr orbit the Coulomb attraction provides the centripetal force:

   $\frac{k e^2}{r^2} = \frac{m v^2}{r} \;\Rightarrow\; m v^2 = \frac{k e^2}{r}.$

   So the kinetic energy is

   $K = \tfrac{1}{2} m v^2 = \frac{k e^2}{2r}.$ …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is

   (a) 9R

   (b) R/3

   (c) 3R

   (d) R/9

   ›Reveal solutionSolution

   Bohr radius of the nth orbit scales as $n^2$, so the third orbit's radius is $9R$.

   In the Bohr model, the radius of the $n$th orbit of the hydrogen atom is

   $r_n = n^2 r_1$

   …

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10. CBSE 2026Set ANNUAL1 mark
    Q.Write the definition of emission line spectrum.

    ›Reveal solutionSolution

    When atoms of a rarefied gas are excited (e.g. by heating or an electric discharge) and their electrons fall back to lower energy levels, they emit light only at specific wavelengths, producing a spectrum of separated bright lines rather than a continuous band.

    An emission line spectrum is obtained when the light emitted directly by a source of excited atoms (such as a gas discharge tube) is passed through a spectrometer/prism. Because each element's electrons can only occupy discrete (quantised) energy levels, transitions between these levels emit photons of only certain specific energies (and hence specific wavelengths/frequencies). The resulting spectrum consists of a series of bright, sharp coloured lines at these part …

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11. CBSE 2026Set ANNUAL1 markMCQ
    Q.Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n ∝ n², and the total energy is E_n = −13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i − E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. According to Bohr's second postulate, which quantity is quantized?

    (a) Energy of electron

    (b) Orbital angular momentum

    (c) Linear momentum

    (d) Frequency of revolution

    ›Reveal solutionSolution

    Bohr's second postulate quantizes the electron's orbital angular momentum in integer multiples of $h/2\pi$.

    Bohr's second postulate states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $\hbar = h/2\pi$: …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n ∝ n², and the total energy is E_n = −13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i − E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. The energy of an electron in the first excited state (n = 2) of hydrogen is:

    (a) –13.6 eV

    (b) –6.8 eV

    (c) –3.4 eV

    (d) –1.7 eV

    ›Reveal solutionSolution

    Bohr energy levels of hydrogen follow $E_n = -13.6/n^2$ eV; for $n=2$, $E_2 = -3.4$ eV.

    $E_n = \dfrac{-13.6}{n^2}\,\text{eV}$

    For the first excited state, $n=2$: …

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