Q.v21 and v31 are respectively the frequencies of the first and second lines of the Lyman series in hydrogen spectrum. If v32 is the frequency of the first line of Balmer series, establish the relation among v31, v32 and v21.

Showing the 12 most recent of 70 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, 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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. Transitions ending at n = 3 belong to which series?

   (a) Lyman series

   (b) Balmer series

   (c) Paschen series

   (d) Brackett series

   ›Reveal solutionSolution

   Transitions of hydrogen electrons ending on the $n=3$ level constitute the Paschen series (infrared).

   …

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10. CBSE 2026Set ANNUAL1 mark
    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. Explain the physical significance of the negative sign in the expression for the total energy E_n of an electron in the Bohr model. OR What is the ratio of the radius of the third Bohr orbit (n = 3) to the radius of the first Bohr orbit (n = 1)?

    ›Reveal solutionSolution

    A negative total energy means the electron is in a bound state — energy must be added to remove it to infinity (where E=0).

    The total energy of an electron in the $n^{th}$ Bohr orbit is $E_n = -\dfrac{13.6}{n^2}\,\text{eV}$. The negative sign signifies that the electron is bound to the nucleus by the attractive electrostatic force — its potential energy (taken as zero at infinite separation) is more negative than its kinetic energy is positive, so the net energy is negative. This means external energy (at least $|E_n|$) must be supplied …

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11. CBSE 2026Set ANNUAL1 markMCQ
    Q.In which part of the electromagnetic spectrum, is the Lyman series of hydrogen found?

    (a) Ultraviolet

    (b) Scene

    (c) Infrared

    (d) X-rays

    ›Reveal solutionSolution

    The Lyman series corresponds to electron transitions from higher levels $n=2,3,4,\ldots$ down to $n=1$; these are the highest-energy (shortest-wavelength) hydrogen transitions and fall in the ultraviolet.

    Using the Rydberg formula $\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^2}-\dfrac{1}{n^2}\right)$ for $n=2,3,4,\ldots$, the Lyman series wavelengths work out to roughly 91–122 nm, which lies well below the visible range (about 400–700 nm) and squarely in the ultraviolet band. (By contrast, the Ba …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?

    (a) Its kinetic energy increases and its potential and total energies decrease.

    (b) Its kinetic energy decreases, potential energy increases and its total energy remains the same.

    (c) Its kinetic and total energies decrease and its potential energy increases.

    (d) Its kinetic energy, potential energy and total energy all decrease.

    ›Reveal solutionSolution

    In the Bohr model $KE=+\dfrac{ke^2}{2r}$, $PE=-\dfrac{ke^2}{r}=-2KE$, and $E=KE+PE=-\dfrac{ke^2}{2r}$. As the electron moves to a lower (ground) state, $r$ decreases, so $KE$ increases while $PE$ and $E$ both decrease (become more negative).

    Energies in the Bohr model

    For an electron in a circular orbit of radius $r$ about the nucleus (charge $+e$), the Coulomb force supplies the centripetal force:

    $\frac{mv^2}{r}=\frac{ke^2}{r^2}\ \implies\ mv^2=\frac{ke^2}{r}$

    Kinetic energy:

    $KE=\frac12 mv^2=\frac{ke^2}{2r}\quad(\text{always positive})$

    Potential energy (electrostatic PE between electron and nucleus, taking $PE=0$ at $r=\infty$):

    $PE=-\frac{ke^2}{r}\quad(\text{always negative})$

    Total energy:

    $E=KE+PE=\frac{ke^2}{2r}-\frac{ke^2}{r}=-\frac{ke^2}{2r}\quad(\text{always negative})$

    Note that $PE=-2\,KE$ and $E=-KE$ — standard virial-theorem relations for the Coulomb (inverse-square) force.

    Effect of a transition from an excited state to the ground state

    Going from an excited state to the ground state means the electron moves to a smaller orbit radius $r$ (ground state, $n=1$, has the smallest Bohr radius).

    …

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