Q.If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when

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1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A hydrogen atom absorbs energy and rises to $n=3$ state from its ground state $n=1$. If the potential energy of the atom at its ground state is -13.6 eV , find the wave length emitted by it when it returns to its ground state: {Planck's constant $=6.6 \times 10^{34} \mathrm{~J} \mathrm{~s}$ } (A) $12000^{\circ} \mathrm{A}$ (B) $1020^{\circ} \mathrm{A}$ (C) $7000^{\circ} \mathrm{A}$ (D) $4000^{\circ} \mathrm{A}$

   ›Reveal solutionSolution

   The key idea is that the energy of a hydrogen atom in state $n$ is $E_n = -13.6/n^2$ eV. The photon emitted in the transition $n=3 \to n=1$ carries the energy difference, and its wavelength is found via $E = hc/\lambda$. The computed wavelength is about $1020 \, \text{Å}$, matching option (B).

   Concept & Intuition

   In hydrogen, the total energy of an electron in a given orbit is the sum of its kinetic and potential energies. For the ground state ($n=1$), the potential energy is given as $-13.6$ eV, but the total energy is half that: $E_1 = -13.6$ eV (since for Coulomb forces, total energy = potential energy/2). The energy levels scale as $1/n^2$. When the atom drops from a higher level to a lower one, the energy difference is released as a photon. The wavelength of that photon is inversely proportional to the energy difference.

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   Step-by-step solution

   1. Determine the total energy of the ground state The potential energy in the ground state is given as $-13.6$ eV. For a hydrogen atom, the total energy $E_n$ is half the potential energy (virial theorem). So:

   $E_1 = \frac{1}{2} \times (-13.6) = -13.6 \ \text{eV}$

   (This is the standard ground-state energy of hydrogen.)

   2. Find the energy of the $n=3$ state Energy levels scale as $E_n = E_1 / n^2$:

   $E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.511 \ \text{eV}$

   3. Compute the energy of the emitted photon The transition is from $n=3$ to $n=1$:

   $\Delta E = E_3 - E_1 = (-1.511) - (-13.6) = 12.089 \ \text{eV}$

   (We can keep it exact: $\Delta E = 13.6 \left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} = \frac{108.8}{9} \ \text{eV}$.)

   4. Convert energy to joules $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$, so:

   $\Delta E = \frac{108.8}{9} \times 1.6 \times 10^{-19} \approx 12.089 \times 1.6 \times 10^{-19} = 1.934 \times 10^{-18} \ \text{J}$

   5. Use the photon energy–wavelength relation

   $E = \frac{hc}{\lambda} \quad \Rightarrow \quad \lambda = \frac{hc}{E}$

   With $h = 6.6 \times 10^{-34} \ \text{J·s}$ and $c = 3 \times 10^8 \ \text{m/s}$:

   $$$$

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.An atom has a single electron. Its ground state energy is -30 eV and its first excited state energy is -8 eV . The atom is bombarded with a stream of photons, each of energy 15 eV . Assuming the atom being in the ground state, which of the following statements is correct? A. Atom gets excited to the first excited state and later emit photons of 22 eV B. Atom absorbs energy continuously until 22 eV is accumulated and then gets excited C. Atom will not get excited, and the transmitted light will have the same frequency as the incident light D. Atom will absorb the photon and re-emit a photon of lower energy (A) A (B) B (C) D (D) C

   ›Reveal solutionSolution

   The atom can only absorb photons whose energy exactly matches the difference between two allowed energy levels. Since 15 eV does not match the 22 eV gap from ground to first excited state, no absorption occurs, so the light passes through unchanged. The correct option is (D).

   The key concept here is quantized energy levels in an atom. An electron can only jump between discrete energy states by absorbing or emitting a photon whose energy equals the exact difference between those states. If the photon energy doesn't match any allowed transition, the atom simply cannot absorb it — it's like trying to climb a staircase with steps of uneven height using a single leap that lands between steps.

   Let’s work through the reasoning:

   1. Identify the energy levels given Ground state: $E_0 = -30 \, \text{eV}$ First excited state: $E_1 = -8 \, \text{eV}$ The energy difference between them is:

   $\Delta E = E_1 - E_0 = (-8) - (-30) = 22 \, \text{eV}$

   2. Compare the photon energy to the required transition energy

      The incident photons have energy $E_{\text{photon}} = 15 \, \text{eV}$.

      For the atom to absorb a photon and jump from ground to first excited state, the photon must have exactly 22 eV. Since 15 eV ≠ 22 eV, this transition is impossible.

   3. Consider other possible transitions

      Could the atom absorb 15 eV and jump to some other state? The problem only gives two states, so no other bound state is available. The atom cannot absorb a fraction of a photon’s energy, nor can it "store" partial energy from multiple photons (that would violate energy quantization in a single quantum event).

   4. What actually happens?

      Since no allowed transition matches 15 eV, the atom remains in the ground state. The photons are not absorbed; they simply pass through the atom. Thus, the transmitted light has the same frequency (and energy) as the incident light.

   5. Evaluate the options

      • A: Incorrect — the atom cannot get excited to the first excited state with 15 eV. …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.An atom with one electron has ionization energy of 24 eV . An electron in this atom makes a transition from an excited energy level, where $\mathrm{E}=-15 \mathrm{eV}$, to the ground state. What is the wavelength of the emitted photon from this transition? (A) 112 nm (B) 496 nm (C) 775 nm (D) 138 nm

   ›Reveal solutionSolution

   The ionization energy gives the ground-state energy; using the given excited-state energy, the photon energy is the difference, and the wavelength is found via $E = hc/\lambda$. The result is about 138 nm, matching option (D).

   The key concept here is that the ionization energy of a hydrogen-like atom is the energy required to remove the electron from the ground state to infinity (where energy is zero). So if the ionization energy is 24 eV, that means the ground-state energy $E_1$ is $-24$ eV (since we define zero at infinity). The problem gives an excited state at $-15$ eV. When the electron drops from that excited state to the ground state, the emitted photon’s energy equals the difference in energy levels. Then we use the Planck-Einstein relation to find the wavelength.

   Let’s work through it step by step.

   1. Determine the ground-state energy. Ionization energy = 24 eV means it takes 24 eV to remove the electron from the ground state. Therefore, the ground-state energy is

   $E_{\text{ground}} = -24 \text{ eV}.$

   2. Identify the excited-state energy. The problem states the electron is in an excited level with

   $E_{\text{excited}} = -15 \text{ eV}.$

   3. Find the energy of the emitted photon. The transition is from the excited state to the ground state, so the photon energy is the difference:

   $\Delta E = E_{\text{excited}} - E_{\text{ground}} = (-15) - (-24) = 9 \text{ eV}.$

   4. Convert photon energy to wavelength. The relation is $E = \frac{hc}{\lambda},$ …

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4. KCET 2026Set C21 markMCQ
   Q.The radius of first orbit in hydrogen atom is $5.3 \times 10^{-11}$ m. The kinetic energy $E_K$, potential energy $E_P$ and total energy $E_T$ of electron in first orbit are (A) $E_K$ = - 13.6 eV, $E_P$ = 27.2 eV, $E_T$ = 13.6 eV (B) $E_K$ = 13.6 eV, $E_P$ = - 27.2 eV, $E_T$ = - 13.6 eV (C) $E_K$ = - 27.2 eV, $E_P$ = - 13.6 eV, $E_T$ = 13.6 eV (D) $E_K$ = 13.6 eV, $E_P$ = - 6.8 eV, $E_T$ = - 13.6 eV

   ›Reveal solutionSolution

   For an electron in a Coulomb (Bohr) orbit, kinetic energy is the negative of the total energy and potential energy is twice the total energy; hydrogen's first (ground) orbit has total energy $-13.6\ \text{eV}$.

   Step 1 — Recall the Bohr-model energy relations

   For an electron bound in a circular Coulomb orbit:

   $E_K = -E_T,\qquad E_P = 2E_T,\qquad E_T = E_K + E_P$

   For hydrogen's first orbit ($n=1$): $E_T = -13.6\ \text{eV}$ (this is the standard ground-state energy, consistent with the given first-orbit radius $5.3\times10^{-11}\ \text{m}$).

   Step 2 — Compute each energy

   $E_K = -E_T = -(-13.6) = 13.6\ \text{eV}$ …

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5. KCET 2026Set C21 markMCQ
   Q.An electron transition takes place from excited state to ground state in hydrogen atom, then (A) Its kinetic energy increases but potential energy and total energy decrease (B) Its kinetic energy, potential energy and total energy decrease (C) Kinetic energy decreases, potential energy increases but total energy remains same (D) Kinetic energy and total energy decrease but potential energy increases

   ›Reveal solutionSolution

   As an electron falls from an excited orbit to the ground state, its orbit radius shrinks and total energy becomes more negative; using $E_K=-E_T$ and $E_P=2E_T$ shows kinetic energy rises while potential and total energy both fall.

   Step 1 — How total energy changes with orbit

   For the Bohr model, $E_T = -\dfrac{13.6}{n^2}\ \text{eV}$. As the electron transitions from an excited state (larger $n$) down to the ground state ($n=1$), $n$ decreases, so $E_T$ becomes more negative (i.e. total energy decreases).

   Step 2 — Track kinetic energy

   $E_K = -E_T$

   As $E_T$ becomes more negative, $-E_T$ becomes more positive — so kinetic energy increases as the electron falls to a lower orbit (it speeds up, consistent with a smaller orbital radius).

   Step 3 — Track potential energy

   $E_P = 2E_T$ …

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6. COMEDK 2025Set 2025-E1 markMCQ
   Q.The minimum energy required by a hydrogen atom in ground state to emit radiation in Paschen series is nearly: (A) 13.6 eV (B) 12.75 eV (C) 10.75 eV (D) 1.5 eV

   ›Reveal solutionSolution

   The key is that the Paschen series involves transitions to $n=3$, so the minimum energy to emit such radiation from the ground state requires exciting the atom to $n=4$ first. That energy difference is $13.6\,\text{eV} \left(\frac{1}{1^2} - \frac{1}{4^2}\right) = 12.75\,\text{eV}$, so the answer is (B).

   The Paschen series corresponds to all transitions where an electron falls to the $n=3$ energy level. For a hydrogen atom initially in the ground state ($n=1$) to emit any Paschen-series photon, it must first be excited to a higher level from which it can then drop to $n=3$. The minimum energy required is the smallest excitation that allows a subsequent Paschen emission — that is, exciting the atom from $n=1$ to $n=4$. (Why $n=4$? Because from $n=4$ the electron can fall directly to $n=3$; from $n=2$ or $n=3$ it cannot emit a Paschen photon, since those are below or at the final level.)

   1. Recall the hydrogen energy levels The energy of a level with principal quantum number $n$ is

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

   For the ground state, $E_1 = -13.6\,\text{eV}$.

   2. Identify the target level for the smallest Paschen emission

      The smallest energy photon in the Paschen series comes from the transition $n=4 \to n=3$. So the atom must first reach $n=4$.

   3. Compute the excitation energy from $n=1$ to $n=4$

   $\Delta E = E_4 - E_1 = \left(-\frac{13.6}{16}\right) - \left(-13.6\right) = 13.6\left(1 - \frac{1}{16}\right) = 13.6 \times \frac{15}{16}.$

   $\Delta E = \frac{13.6 \times 15}{16} = \frac{204}{16} = 12.75\,\text{eV}.$

   4. Interpret the result …

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7. COMEDK 2025Set 2025-M1 markMCQ
   Q.Which, of the following is true of the Balmer series of the hydrogen spectrum? a. The series is in the visible region. b. The entire series falls in the ultraviolet region c. The entire series falls in the infrared region d. The series is partly in the visible region and partly in the infrared region (A) b (B) c (C) a (D) d

   ›Reveal solutionSolution

   The Balmer series (transitions ending at $n = 2$) is the visible-light series of hydrogen — its lines lie in the visible region. The correct statement is a, option (C).

   Concept

   The Balmer series is produced when an electron falls from a level $n \ge 3$ to $n = 2$. Historically it is known as the visible series of hydrogen because its bright lines (red, blue-green, violet) fall in the visible part of the spectrum. The Rydberg formula lets us confirm the wavelengths.

   Solution

   1. Rydberg formula with $n_f = 2$:

   $\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{n^2}\right),\quad n = 3, 4, 5, \dots$

   2. First line ($n = 3 \to 2$): $\lambda \approx 656\,\text{nm}$ (red).
   3. Further lines: $486\,\text{nm}$ (blue-green), $434\,\text{nm}$ and $410\,\text{nm}$ (violet) — all visible. …

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8. COMEDK 2024Set 2024-A1 markMCQ
   Q.The shortest wavelengths of Paschen, Lymen and Balmer series are in the ratio (A) $9: 1: 4$ (B) $4: 1: 9$ (C) $2:1:3$ (D) $3: 1: 2$

   ›Reveal solutionSolution

   The shortest wavelength in a spectral series corresponds to the transition from $n=\infty$ to the lowest level $n_1$ of that series, so $\lambda_{\min}\propto n_1^2$. For Paschen ($n_1=3$), Lyman ($n_1=1$) and Balmer ($n_1=2$) the shortest wavelengths are in the ratio $9:1:4$. The correct option is (A).

   ---

   The key idea is that the shortest wavelength in any spectral series occurs when the electron jumps from the highest possible energy level (effectively $n = \infty$) down to the series' ground level. The wavelength is then inversely proportional to the square of the lowest level's principal quantum number.

   Why this works

   For hydrogen-like atoms, the Rydberg formula gives the wavenumber (inverse wavelength) for a transition from $n_2$ to $n_1$:

   $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

   where $R$ is the Rydberg constant.

   The shortest wavelength corresponds to the largest energy difference, which happens when $n_2 \to \infty$. Then:

   $\frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{n_1^2} - 0 \right) = \frac{R}{n_1^2}$

   So $\lambda_{\text{min}} \propto n_1^2$.

   Thus, the ratio of shortest wavelengths for different series is simply the ratio of the squares of their respective lowest principal quantum numbers.

   Step-by-step

   1. Identify the lowest level for each series

      • Lyman series: transitions end at $n_1 = 1$
      • Balmer series: transitions end at $n_1 = 2$
      • Paschen series: transitions end at $n_1 = 3$

   2. Write the shortest wavelength for each

   $\lambda_{\text{Lyman}} \propto 1^2 = 1$

   $\lambda_{\text{Balmer}} \propto 2^2 = 4$

   $\lambda_{\text{Paschen}} \propto 3^2 = 9$

   3. Form the ratio (Paschen : Lyman : Balmer) $$$$

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9. COMEDK 2024Set 2024-E1 markMCQ
   Q.If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength $\lambda$. When it jumps from the second to the first orbit, the corresponding wavelength of the photon will be (A) $\frac{5 \lambda}{27}$ (B) $\frac{7 \lambda}{20}$ (C) $\frac{16 \lambda}{9}$ (D) $\frac{20 \lambda}{7}$

   ›Reveal solutionSolution

   Photon energy $\propto\left(\tfrac{1}{n_f^2}-\tfrac{1}{n_i^2}\right)$ and $\lambda\propto 1/E$; the ratio of the $2\to1$ to $3\to2$ energies gives $\lambda_{2\to1}=\tfrac{5\lambda}{27}$.

   Using the Rydberg relation, the emitted photon's wavenumber (and hence energy) is

   $\frac{1}{\lambda} \propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).$

   Transition $3\to2$ (wavelength $\lambda$):

   $\frac{1}{\lambda} \propto \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}.$

   Transition $2\to1$ (wavelength $\lambda'$):

   $\frac{1}{\lambda'} \propto \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}.$ …

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10. KCET 2023Set A-31 markMCQ
    Q.Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then. (A) $\lambda_1 = \dfrac{\lambda_2 \lambda_3}{\lambda_2 + \lambda_3}$ (B) $\lambda_2 = \lambda_1 + \lambda_3$ (C) $\lambda_2 = \dfrac{\lambda_1 \lambda_3}{\lambda_1 + \lambda_3}$ (D) $\lambda_3 = \dfrac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$

    ›Reveal solutionSolution

    Energy — not wavelength — is additive, so write $E = hc/\lambda$ for each transition and add the two-step energies to get the one-step energy.

    1. Read the transitions off the diagram

    • $\lambda_1$: $E_2 \to E_1$ ⇒ photon energy $\dfrac{hc}{\lambda_1} = E_2 - E_1$
    • $\lambda_2$: $E_3 \to E_1$ ⇒ photon energy $\dfrac{hc}{\lambda_2} = E_3 - E_1$
    • $\lambda_3$: $E_3 \to E_2$ ⇒ photon energy $\dfrac{hc}{\lambda_3} = E_3 - E_2$

    2. The energy identity

    The level energies simply telescope:

    $(E_3 - E_1) = (E_3 - E_2) + (E_2 - E_1)$

    This is the physical heart of the problem: whether the atom drops from $E_3$ to $E_1$ in one jump or via $E_2$, the total energy released is the same (conservation of energy).

    3. Convert to wavelengths

    $\frac{hc}{\lambda_2} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_1}$

    Cancel $hc$:

    $\frac{1}{\lambda_2} = \frac{1}{\lambda_1} + \frac{1}{\lambda_3} = \frac{\lambda_3 + \lambda_1}{\lambda_1 \lambda_3}$

    $\boxed{\;\lambda_2 = \frac{\lambda_1 \lambda_3}{\lambda_1 + \lambda_3}\;}$

    4. Why the other options fail …

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11. COMEDK 2023Set 2023-E1 markMCQ
    Q.The ground state energy of hydrogen atom is $-13.6 \mathrm{~eV}$. If the electron jumps from the $3^{\text {rd }}$ excited state to the ground state then the energy of the radiation emitted will be: (A) 1.275 MeV (B) 12.75 eV (C) 12.75 J (D) 12.75 MeV

    ›Reveal solutionSolution

    The unit must be eV (a few eV is the atomic scale; MeV is a nuclear scale, and 12.75 J is absurdly large for one photon).

    Concept: Bohr energy levels of hydrogen, E_n = -13.6/n^2 eV. The '3rd excited state' is n = 4 (ground n = 1, 1st excited n = 2, 2nd excited n = 3, 3rd excited n = 4).

    E_4 = -13.6 / 16 = -0.85 eV

    E_1 = -13.6 eV

    Energy of the emitted photon:

    E = E_4 - E_1 = -0.85 - (-13.6) = 12.75 eV. …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.The wavelength of the first line of Lyman series for $\mathrm{H}$ - atom is equal to that of the second line of Balmer series for a $\mathrm{H}$-like ion. The atomic number $\mathrm{Z}$ of $\mathrm{H}$-like ion is (A) 4 (B) 1 (C) 2 (D) 3

    ›Reveal solutionSolution

    Equating the first Lyman line of H ($3R/4$) to the second Balmer line ($n=4\to2$) of a hydrogen-like ion of charge $Z$ gives $3R/4 = 3RZ^2/16$, so $Z^2=4$, $Z=2$.

    First line of Lyman series for H ($n=2\to1$):

    $\frac{1}{\lambda_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3R}{4}.$

    Second line of Balmer series ($n=4\to2$) for a H-like ion of atomic number $Z$: …

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