Q.What is the total number of orbitals associated with the principal quantum number n = 3?

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The frequency of photon which is emitted during a transition of electron of $\mathrm{He}^{+}$ion from fifth energy level to third energy level will be: (A) $1.34 \times 10^{-14} \mathrm{~s}^{-1}$ (B) $9.39 \times 10^{14} \mathrm{~s}^{-1}$ (C) $2.34 \times 10^{14} \mathrm{~s}^{-1}$ (D) $8.29 \times 10^{-14} \mathrm{~s}^{-1}$

   ›Reveal solutionSolution

   For a hydrogen-like ion, the transition energy is given by $\Delta E = R_H Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, and the photon frequency is $\nu = \Delta E / h$. For $\mathrm{He}^{+}$ ($Z=2$) from $n=5$ to $n=3$, the frequency is $9.39 \times 10^{14} \, \mathrm{s}^{-1}$, which corresponds to option (B).

   The key concept here is the Bohr model for hydrogen-like ions (ions with only one electron, like $\mathrm{He}^{+}$). The energy levels depend on the square of the atomic number $Z$, so the transition energy (and thus the photon frequency) scales with $Z^2$. The problem asks for the frequency, not the wavelength, so we directly use $\nu = \Delta E / h$.

   1. Recall the energy level formula For a hydrogen-like atom, the energy of the $n$-th level is:

   $E_n = - \frac{R_H Z^2}{n^2}$

   where $R_H = 2.18 \times 10^{-18} \, \mathrm{J}$ is the Rydberg constant (in energy units), and $Z=2$ for $\mathrm{He}^{+}$.

   2. Find the energy difference The transition is from $n_i = 5$ to $n_f = 3$:

   $\Delta E = E_3 - E_5 = R_H Z^2 \left( \frac{1}{3^2} - \frac{1}{5^2} \right)$

   Compute the bracket:

   $\frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225}$

   So:

   $\Delta E = (2.18 \times 10^{-18}) \times (4) \times \frac{16}{225}$

   3. Calculate numerically First, $4 \times \frac{16}{225} = \frac{64}{225} \approx 0.28444$. Then:

   $\Delta E = 2.18 \times 10^{-18} \times 0.28444 \approx 6.20 \times 10^{-19} \, \mathrm{J}$

   (More precisely: $2.18 \times 10^{-18} \times 64 / 225 = (139.52 / 225) \times 10^{-18} = 0.62009 \times 10^{-18} = 6.2009 \times 10^{-19} \, \mathrm{J}$.)

   4. Convert energy to frequency Use Planck’s relation $\Delta E = h \nu$, with $h = 6.626 \times 10^{-34} \, \mathrm{J \cdot s}$:

   $\nu = \frac{\Delta E}{h} = \frac{6.20 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 9.36 \times 10^{14} \, \mathrm{s}^{-1}$

   A more precise calculation gives $9.39 \times 10^{14} \, \mathrm{s}^{-1}$, matching option (B).

   Tip

   A quick check: the Rydberg formula for wavenumber is $\tilde{\nu} = R_\infty Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, and frequency $\nu = c \tilde{\nu}$. Using $R_\infty = 1.097 \times 10^7 \, \mathrm{m}^{-1}$ and $c = 3 \times 10^8 \, \mathrm{m/s}$ gives the same result — a good cross-check.

   Watch out

   A common mistake is to forget the $Z^2$ factor or to use $Z=1$ (as for hydrogen). Also, note that the options include two with negative exponents ($10^{-14}$) — those are clearly too small for a visible/UV photon frequency, so they can be eliminated immediately.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.The Bohr orbit of the hydrogen atom ( $n=1$ ) is $0.530 \mathop {\rm{A}}\limits^{\rm{o}}$. The radius of the first excited state orbit in $\mathop {\rm{A}}\limits^{\rm{o}}$ is: (A) 0.16 (B) 5.21 (C) 1.66 (D) 2.12

   ›Reveal solutionSolution

   The radius of a Bohr orbit scales as $n^2$, so the first excited state ($n=2$) has a radius four times the ground-state radius, giving $4 \times 0.530 = 2.12$ Å. The correct option is (D).

   The key idea is that in the Bohr model of the hydrogen atom, the radius of the $n$-th orbit is proportional to $n^2$. The ground state ($n=1$) radius is given; the first excited state corresponds to $n=2$. So we simply multiply by $2^2 = 4$.

   1. Recall the Bohr radius formula The radius of the $n$-th orbit in hydrogen is

   $r_n = n^2 a_0$

   where $a_0$ is the Bohr radius (the radius for $n=1$). This comes from balancing the Coulomb force and centripetal force, with quantized angular momentum.

   2. Identify the given and required states

      • Ground state: $n=1$, $r_1 = 0.530$ Å.
      • First excited state: $n=2$ (since the first excited state is the next energy level above the ground state).

   3. Apply the scaling

   $r_2 = 2^2 \times r_1 = 4 \times 0.530 = 2.12 \text{ Å}$

   4. Match with the options The value 2.12 Å corresponds to option (D).

   Watch out

   A common mistake is to think the first excited state is $n=1$ (it isn’t — that’s the ground state) or to use $n=3$ (the second excited state). Always remember: ground state = $n=1$, first excited = $n=2$, second excited = $n=3$, etc.

   Tip

   You don’t need to memorize the Bohr radius value; just remember the $n^2$ scaling. If the ground-state radius is given, any other orbit’s radius is just a square-multiple away.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. COMEDK 2025Set 2025-E1 markMCQ
   Q.The ratio of the difference between the radii of $3^{\text {rd }}$ and $4^{\text {th }}$ orbits of the $\mathrm{He}^{+}$and those of $\mathrm{Li}^{2+}$ is: (A) $2: 3$ (B) $3: 1$ (C) $1: 3$ (D) $3: 2$

   ›Reveal solutionSolution

   The key idea is that the Bohr radius for a hydrogen-like ion scales as $n^2/Z$. Computing the difference $r_4 - r_3$ for He⁺ and Li²⁺ and taking the ratio gives $3:2$, so the answer is (D).

   Concept & Intuition

   In the Bohr model, the radius of the $n$-th orbit for a hydrogen-like ion (one electron, nuclear charge $Ze$) is

   $r_n = \frac{n^2 a_0}{Z},$

   where $a_0$ is the Bohr radius (a constant). The difference between two orbits depends only on the change in $n^2$ divided by $Z$. For He⁺, $Z=2$; for Li²⁺, $Z=3$. The ratio of the differences $(r_4 - r_3)$ for the two ions simplifies to a simple fraction of their $Z$ values, because the $n^2$ part cancels.

   Step-by-step solution

   1. Write the general formula For any hydrogen-like ion:

   $r_n = \frac{n^2 a_0}{Z}.$

   Here $a_0$ is the same for all ions.

   2. Compute the difference for He⁺ ($Z=2$)

   $r_4 - r_3 = \frac{4^2 a_0}{2} - \frac{3^2 a_0}{2} = \frac{16 a_0}{2} - \frac{9 a_0}{2} = \frac{7 a_0}{2}.$

   3. Compute the difference for Li²⁺ ($Z=3$)

   $r_4 - r_3 = \frac{4^2 a_0}{3} - \frac{3^2 a_0}{3} = \frac{16 a_0}{3} - \frac{9 a_0}{3} = \frac{7 a_0}{3}.$

   4. Take the ratio

   $\frac{\text{He}^+ \text{ difference}}{\text{Li}^{2+} \text{ difference}} = \frac{7 a_0 / 2}{7 a_0 / 3} = \frac{1/2}{1/3} = \frac{3}{2}.$

   So the ratio is $3:2$.

   Tip

   Notice the $7a_0$ cancels immediately. The ratio is simply the inverse ratio of the nuclear charges: $\frac{1/Z_{\text{He}}}{1/Z_{\text{Li}}} = \frac{Z_{\text{Li}}}{Z_{\text{He}}} = \frac{3}{2}$. This shortcut works because the $n^2$ difference is the same for both ions.

   Watch out

   A common mistake is to forget that the radius formula uses $n^2$, not $n$. Also, be careful to subtract in the correct order — the problem asks for the difference between the 4th and 3rd orbits, not the other way around.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. COMEDK 2024Set 2024-A1 markMCQ
   Q.The ratio of the first three radii of Bohr's atom is: (A) $1: 4: 9$ (B) $1: 5: 9$ (C) $1:4:27$ (D) $1: 2: 3$

   ›Reveal solutionSolution

   The radii of Bohr orbits scale as $n^2$, so the first three radii are in the ratio $1^2 : 2^2 : 3^2 = 1 : 4 : 9$. The correct option is (A).

   The key concept here is the Bohr model of the atom, which quantizes the angular momentum of an electron orbiting the nucleus. From this quantization, the radius of the $n$-th orbit is derived to be proportional to $n^2$, where $n$ is the principal quantum number (1, 2, 3, …). This is a direct consequence of balancing the Coulomb force with the centripetal force and imposing angular momentum quantization.

   Why this works:

   Instead of memorizing the formula, think of it this way: the electron’s orbit is stable only when its angular momentum is an integer multiple of $h/2\pi$. This forces the radius to grow as the square of the orbit number. So the first three radii are simply in the ratio of the squares of 1, 2, and 3.

   Let’s walk through the derivation step by step.

   1. Start with the Bohr quantization condition The angular momentum of the electron in the $n$-th orbit is

   $m v r_n = n \frac{h}{2\pi}$

   where $m$ is the electron mass, $v$ its speed, $r_n$ the radius of the $n$-th orbit, and $h$ Planck’s constant.

   2. Apply the Coulomb force as the centripetal force The electrostatic attraction between the nucleus (charge $+Ze$) and the electron (charge $-e$) provides the centripetal force:

   $\frac{k e^2}{r_n^2} = \frac{m v^2}{r_n}$

   where $k = 1/(4\pi\epsilon_0)$. For hydrogen, $Z=1$.

   3. Eliminate $v$ between the two equations From step 1, $v = \frac{n h}{2\pi m r_n}$. Substitute into step 2:

   $\frac{k e^2}{r_n^2} = \frac{m}{r_n} \left( \frac{n h}{2\pi m r_n} \right)^2$

   Simplify:

   $\frac{k e^2}{r_n^2} = \frac{n^2 h^2}{4\pi^2 m r_n^3}$

   4. Solve for $r_n$ Multiply both sides by $r_n^3$:

   $k e^2 r_n = \frac{n^2 h^2}{4\pi^2 m}$

   Hence:

   $r_n = \frac{n^2 h^2}{4\pi^2 m k e^2}$

   All constants are fixed, so $r_n \propto n^2$.

   5. Write the ratio for $n=1,2,3$

   $r_1 : r_2 : r_3 = 1^2 : 2^2 : 3^2 = 1 : 4 : 9$

   Watch out

   A common mistake is to think the radii are proportional to $n$ (like the energy levels, which go as $1/n^2$). Remember: radius grows with $n^2$, not $n$.

   Tip

   You can remember this as “Bohr radii go like the squares of the orbit numbers” — a quick mental check: the first radius is the Bohr radius $a_0$, the second is $4a_0$, the third is $9a_0$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. COMEDK 2024Set 2024-E1 markMCQ
   Q.What is the wave number (Units $\mathrm{cm}^{-1}$) of the longest wave length transition in the Balmer series of Hydrogen spectrum? $Z=1 \text { for } H$ (A) 27419.6 (B) 15233 (C) 39605.6 (D) 1354

   ›Reveal solutionSolution

   The longest wavelength transition in the Balmer series corresponds to the smallest energy jump within that series, which is from $n=3$ to $n=2$. Using the Rydberg formula, the wave number is $15233\ \mathrm{cm}^{-1}$, so the correct option is (B).

   The Balmer series describes transitions in hydrogen where the electron falls to the $n=2$ energy level. The longest wavelength means the smallest energy photon emitted, which occurs for the smallest possible energy difference — that is, from the nearest higher level, $n=3$ to $n=2$. The wave number $\tilde{\nu}$ (in $\mathrm{cm}^{-1}$) is directly proportional to the energy difference, so we use the Rydberg formula for hydrogen:

   $\tilde{\nu} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

   where $R_H = 109677\ \mathrm{cm}^{-1}$ (the Rydberg constant for hydrogen), $n_f = 2$ (Balmer series), and $n_i = 3$ for the longest wavelength transition.

   1. Identify the transition: For the Balmer series, the final level is $n_f = 2$. The smallest energy jump (longest wavelength) is from the lowest possible initial level above 2, which is $n_i = 3$. Any higher $n_i$ would give a larger energy difference and thus a shorter wavelength.

   2. Plug into the Rydberg formula:

   $\tilde{\nu} = 109677 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$

   Compute the fractions:

   $\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$

   3. Multiply:

   $\tilde{\nu} = 109677 \times \frac{5}{36}$

   First, divide $109677$ by $36$:

   $109677 \div 36 = 3046.58333\ldots$

   Then multiply by 5:

   $3046.58333\ldots \times 5 = 15232.9166\ldots$

   4. Round appropriately: The given options are to one decimal place, so $15233\ \mathrm{cm}^{-1}$ matches option (B).

   Tip

   A common mistake is to pick the shortest wavelength transition (from $n=\infty$ to $n=2$), which gives the largest wave number $27419.6\ \mathrm{cm}^{-1}$ — that’s option (A). Always check whether the question asks for longest or shortest wavelength.

   Watch out

   The Rydberg constant is sometimes given as $1.097 \times 10^7\ \mathrm{m}^{-1}$, but here we need it in $\mathrm{cm}^{-1}$. Remember: $1\ \mathrm{m}^{-1} = 0.01\ \mathrm{cm}^{-1}$, so $1.097 \times 10^7\ \mathrm{m}^{-1} = 109700\ \mathrm{cm}^{-1}$ (approximately). The precise value $109677\ \mathrm{cm}^{-1}$ is standard for hydrogen.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

6. COMEDK 2023Set 2023-E1 markMCQ
   Q.The energy of an electron in the ground state of Hydrogen atom is $-2.18 \times 10^{-18} \mathrm{~J}$. What would be the energy associated with the second excited state of $\mathrm{Li}$ ? (A) $-2.18 \times 10^{-18} \mathrm{~J}$ (B) $-4.905 \times 10^{-18} \mathrm{~J}$ (C) $-0.242 \times 10^{-18} \mathrm{~J}$ (D) $-3.26 \times 10^{-18} \mathrm{~J}$

   ›Reveal solutionSolution

   Using $E_n=-2.18\times10^{-18}(Z^2/n^2)$ J for the one-electron ion $\mathrm{Li^{2+}}$ ($Z=3$) in its second excited state ($n=3$): $Z^2/n^2 = 9/9 = 1$, so $E = -2.18\times10^{-18}$ J.

   The Bohr energy of a hydrogen-like (one-electron) species is

   $E_n = -2.18\times10^{-18}\ \frac{Z^2}{n^2}\ \text{J}$

   Species: "Li" as a hydrogen-like ion means $\mathrm{Li^{2+}}$, so $Z = 3$.

   Second excited state: ground $=n=1$, first excited $=n=2$, second excited $=n=3$.

   $E = -2.18\times10^{-18}\times\frac{3^2}{3^2} = -2.18\times10^{-18}\times 1 = -2.18\times10^{-18}\ \text{J}$

   ✓Final answer

   The correct option is (A) — $-2.18\times10^{-18}\ \mathrm{J}$

7. COMEDK 2023Set 2023-M1 markMCQ
   Q.Bohr's radius of 2 nd orbit of $\mathrm{Be}^{3+}$ is equal to that of (A) 4th orbit of hydrogen (B) 2nd orbit of $\mathrm{He}^{+}$ (C) $3 \mathrm{rd}$ obit of $\mathrm{Li}^{2+}$ (D) 1st orbit of hydrogen

   ›Reveal solutionSolution

   The Bohr radius scales as $r_n \propto n^2/Z$. For $\mathrm{Be^{3+}}$ ($Z=4$), the 2nd orbit gives $n^2/Z = 4/4 = 1$, the same value as hydrogen's 1st orbit ($1/1=1$).

   Bohr radius:

   $r_n = r_0\,\frac{n^2}{Z}, \quad r_0 = 0.529\,\text{Å}.$

   For $\mathrm{Be^{3+}}$ ($Z=4$), 2nd orbit ($n=2$):

   $r = r_0\frac{2^2}{4} = r_0\cdot 1 = r_0.$

   Compare with the options ($n^2/Z$):

   • (A) H, $n=4$: $16/1 = 16$.
   • (B) $\mathrm{He^+}$, $n=2$: $4/2 = 2$.
   • (C) $\mathrm{Li^{2+}}$, $n=3$: $9/3 = 3$.
   • (D) H, $n=1$: $1/1 = 1$. ✓

   The 1st orbit of hydrogen has the same radius $r_0$.

   ✓Final answer

   The correct option is (D) — 1st orbit of hydrogen.

8. COMEDK 2022Set 20221 markMCQ
   Q.The spectrum of He$^+$ is similar to (A) H (B) Li$^+$ (C) Na (D) He$^+$

   ›Reveal solutionSolution

   Checking the options: H (1 electron - hydrogen-like, YES). Li+ has 2 electrons (not one-electron). Na has 11 electrons. Option (D) merely repeats He+.

   Concept: Hydrogen-like (one-electron) species show a spectrum of the same FORM as hydrogen, because the Bohr/Schrodinger treatment applies exactly:

   E_n = -13.6 Z^2 / n^2 eV

   He+ has Z = 2 with only ONE electron, so it is hydrogen-like and its spectrum is similar in structure to that of H (the lines are simply scaled by Z^2 = 4).

   Checking the options: H (1 electron - hydrogen-like, YES). Li+ has 2 electrons (not one-electron). Na has 11 electrons. Option (D) merely repeats He+.

   ✓Final answer

   The correct option is (A) — H

   ANSWER: A

9. COMEDK 2022Set 20221 markMCQ
   Q.The Paschen series of hydrogen spectrum lies in which region? (A) UV region (B) IR region (C) Visible region (D) Microwave region

   ›Reveal solutionSolution

   The Paschen series therefore lies in the infrared (IR) region.

   Concept: Series in the hydrogen spectrum, classified by the lower level n1:

   Lyman (n1 = 1) -> ULTRAVIOLET

   Balmer (n1 = 2) -> VISIBLE

   Paschen (n1 = 3) -> INFRARED

   Brackett(n1 = 4) -> infrared

   Pfund (n1 = 5) -> far infrared

   The Paschen series therefore lies in the infrared (IR) region.

   ✓Final answer

   The correct option is (B) — IR region

   ANSWER: B

10. COMEDK 2021Set 20211 markMCQ
    Q.The total number of nodes are given by (A) $(n + 1)$ (B) $(n -l- 1)$ (C) $(n - 1)$ (D) $(n -l+ 1)$

    ›Reveal solutionSolution

    The total node count depends only on the principal quantum number: n - 1.

    Concept: Nodes in an atomic orbital.

    • Number of radial (spherical) nodes = n - l - 1
    • Number of angular (nodal planes) = l

    Total nodes = (n - l - 1) + l = n - 1.

    The total node count depends only on the principal quantum number: n - 1.

    ✓Final answer

    The correct option is (C) — $(n - 1)$

    ANSWER: C

11. COMEDK 2021Set 20211 markMCQ
    Q.Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region? (A) Balmer series (B) Paschen series (C) Brackett series (D) Lyman series

    ›Reveal solutionSolution

    Only the Balmer series lies in the visible region.

    Concept: Hydrogen spectral series and the region each falls in.

    • Lyman series (n -> 1): ULTRAVIOLET
    • Balmer series (n -> 2): VISIBLE (H-alpha 656 nm red, H-beta 486 nm, H-gamma 434 nm, H-delta 410 nm)
    • Paschen (n -> 3), Brackett (n -> 4), Pfund (n -> 5): INFRARED

    Only the Balmer series lies in the visible region.

    ✓Final answer

    The correct option is (A) — Balmer series

    ANSWER: A

12. COMEDK 2021Set 20211 markMCQ
    Q.The Lyman series of hydrogen spectrum lies in which region (A) Infrared region (B) Visible region (C) Ultraviolet region (D) X-ray region

    ›Reveal solutionSolution

    That wavelength range lies in the ULTRAVIOLET region (roughly 10-400 nm), not the visible (400-700 nm) or the infrared.

    Concept: Hydrogen emission series and their spectral regions.

    The Lyman series consists of transitions from n = 2, 3, 4, ... down to n = 1. These are the largest energy gaps in the hydrogen atom (the n=2 -> n=1 line alone is 10.2 eV, lambda = 121.6 nm), giving the shortest wavelengths of the H spectrum, from about 91 nm to 122 nm.

    That wavelength range lies in the ULTRAVIOLET region (roughly 10-400 nm), not the visible (400-700 nm) or the infrared.

    ✓Final answer

    The correct option is (C) — Ultraviolet region

    ANSWER: C