Q.A sparingly soluble salt having general formula A_x^{p+} B_y^{q-} and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Gold number of a protective colloid (A) is $x$. A mixture is prepared by adding $50$ mL of $10\%$ NaCl solution to $500$ mL of gold sol. What is the minimum mass (in mg) of A to be added to the solution to prevent the coagulation of gold sol? (A) $50x$ (B) $500x$ (C) $5x$ (D) $0.5x$

   ›Reveal solutionSolution

   Gold number is defined for $10\text{ mL}$ sol $+\,1\text{ mL}$ of $10\%$ NaCl; here both sol and NaCl are scaled up $\times 50$, so mass needed $= 50x$ mg.

   The gold number $x$ = milligrams of protective colloid $A$ that just prevents coagulation of $10\text{ mL}$ of standard gold sol on adding $1\text{ mL}$ of $10\%$ NaCl.

   In this problem:

   $\text{gold sol} = 500\text{ mL} = 50 \times 10\text{ mL}, \qquad 10\%\ \text{NaCl} = 50\text{ mL} = 50 \times 1\text{ mL}$ …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.AB crystallizes in a bcc lattice. If the distance between two oppositely charged ions in the lattice is 335 pm, then the edge length of it (in pm) is (A) 376.8 (B) 366.8 (C) 386.8 (D) 396.8

   ›Reveal solutionSolution

   In a body-centered cubic (bcc) lattice of a compound AB, the distance between oppositely charged ions is half the body diagonal, so the edge length is $a = \frac{2 \times 335}{\sqrt{3}} \approx 386.8$ pm, matching option (C).

   Concept & Intuition

   The key is to visualize the geometry of a bcc lattice. In AB crystallizing in bcc, the ions are arranged such that one type (say A) sits at the corners and the other (B) at the body center, or vice versa. The shortest distance between an A and a B ion is along the body diagonal — from a corner to the center of the cube. That distance is exactly half the length of the full body diagonal. The body diagonal of a cube of edge length $a$ is $\sqrt{3}\,a$. So if the given distance (335 pm) is that half-diagonal, we can solve for $a$.

   Step-by-step solution

   1. Identify the relevant distance

      In a bcc lattice, the closest oppositely charged ions are a corner ion and the body-centered ion. The distance between them is half the body diagonal of the cube.

   2. Write the relationship

      Body diagonal of a cube = $\sqrt{3}\,a$.

      Half of that = $\frac{\sqrt{3}}{2}\,a$.

      This is given as 335 pm:

   $\frac{\sqrt{3}}{2}\,a = 335$

   3. Solve for edge length $a$ Multiply both sides by 2:

   $\sqrt{3}\,a = 670$

   Divide by $\sqrt{3}$:

   $$a = \frac{670}{\sqrt{3}} …$$

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3. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.100 mL of 0.05 M Cu$^{2+}$ aqueous solution is added to 1 L of 0.1 M KI solution. The number of moles of I$_2$ and CuI$_2$ formed are respectively (A) $5 \times 10^{-3},\ 2.5 \times 10^{-3}$ (B) $5 \times 10^{-3},\ 5 \times 10^{-3}$ (C) $2.5 \times 10^{-3},\ 2.5 \times 10^{-3}$ (D) $2.5 \times 10^{-3},\ 5 \times 10^{-3}$

   ›Reveal solutionSolution

   2Cu²⁺+4I⁻→Cu₂I₂+I₂; 5×10⁻³ mol Cu²⁺ gives 2.5×10⁻³ mol each of I₂ and the precipitate.

   Cu²⁺ does not form a stable CuI₂; it oxidises I⁻ and is itself reduced to Cu(I), precipitating as Cu₂I₂: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂. n(Cu²⁺)=0.100×0.05=5×10⁻³ mol is limiting (I⁻ = 1×0.1 = 0.1 mol, large excess). For every 2 mol Cu²⁺, 1 mol I₂ and 1 m …

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4. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
   Q.The ratio of radii of second orbits of $\mathrm{Li}^{2+}$ and $\mathrm{Be}^{3+}$ is (A) $4 : 3$ (B) $2 : 3$ (C) $3 : 2$ (D) $3 : 4$

   ›Reveal solutionSolution

   The radius of an electron's orbit in a hydrogen-like atom is directly proportional to the square of the principal quantum number ($n^2$) and inversely proportional to the atomic number ($Z$). For the second orbits of $\mathrm{Li}^{2+}$ and $\mathrm{Be}^{3+}$, the ratio of their radii is $\boxed{4:3}$.

   The problem asks for the ratio of the radii of the second orbits of two hydrogen-like species: $\mathrm{Li}^{2+}$ and $\mathrm{Be}^{3+}$. To solve this, we need to understand how the radius of an electron's orbit is determined in such systems.

   Concept and Intuition

   Bohr's model, while having limitations, accurately describes the energy and radii of orbits for hydrogen and hydrogen-like species (atoms or ions with only one electron). Both $\mathrm{Li}^{2+}$ (Lithium with 3 protons, losing 2 electrons leaves 1 electron) and $\mathrm{Be}^{3+}$ (Beryllium with 4 protons, losing 3 electrons leaves 1 electron) are hydrogen-like species.

   According to Bohr's model, the radius of the $n$-th orbit ($r_n$) for a hydrogen-like atom with atomic number $Z$ is given by a specific formula. This formula shows that the radius depends on two main factors:

   1. The principal quantum number ($n$), which defines the energy level or orbit. Higher $n$ means larger orbits.
   2. The atomic number ($Z$), which represents the number of protons in the nucleus. A higher $Z$ means a stronger attractive force from the nucleus, pulling the electron closer and resulting in smaller orbits.

   The relationship is that $r_n$ is directly proportional to $n^2$ and inversely proportional to $Z$. This proportionality is crucial for calculating ratios, as many constants cancel out.

   The radius of the $n$-th orbit in a hydrogen-like atom is given by:

   $r_n = \frac{\epsilon_0 h^2 n^2}{\pi m e^2 Z} = 0.529 \frac{n^2}{Z} \text{ Å}$

   where $\epsilon_0$ is the permittivity of free space, $h$ is Planck's constant, $m$ is the mass of the electron, $e$ is the elementary charge, $n$ is the principal quantum number, and $Z$ is the atomic number.

   For ratio calculations, we can simply use the proportionality:

   $r_n \propto \frac{n^2}{Z}$

   Let's apply this understanding to find the required ratio.

   Step-by-step Solution

   1. Identify the relevant parameters for $\mathrm{Li}^{2+}$:
      • For Lithium (Li), the atomic number $Z = 3$.
      • The problem specifies the "second orbit", so the principal quantum number $n = 2$.
      • Using the proportionality $r_n \propto \frac{n^2}{Z}$, the radius of the second orbit for $\mathrm{Li}^{2+}$ can be expressed as: …

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5. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.At T (K), $K_c$ value for the reaction $\frac{1}{3} \mathrm{N}_2

   (g) + \mathrm{H}_2

   (g) \rightleftharpoons \frac{2}{3} \mathrm{NH}_3 (g)$is 50. The$K_c$value for the reaction$2\mathrm{NH}_3

   (g) \rightleftharpoons \mathrm{N}_2

   (g) + 3\mathrm{H}_2 (g)$at the same temperature is (A)$4 \times 10^{-6}$(B)$8 \times 10^{-6}$(C)$6 \times 10^{-6}$(D)$8 \times 10^{-3}$

   ›Reveal solutionSolution

   The key idea is to relate equilibrium constants when the reaction is reversed and scaled. Given $K_c = 50$ for $\frac{1}{3}\mathrm{N}_2 + \mathrm{H}_2 \rightleftharpoons \frac{2}{3}\mathrm{NH}_3$, the $K_c$ for $2\mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3\mathrm{H}_2$ is $8 \times 10^{-6}$.

   Concept and Intuition

   Equilibrium constants depend on how the reaction is written. If you multiply a reaction by a factor, the $K_c$ is raised to that power. If you reverse a reaction, $K_c$ becomes its reciprocal. Here, the target reaction is the reverse of the given reaction, but also multiplied by 3 to clear fractions. So we need to apply both transformations.

   Step-by-step solution

   1. Write the given reaction and its $K_c$ Given:

   $\frac{1}{3}\mathrm{N}_2(g) + \mathrm{H}_2(g) \rightleftharpoons \frac{2}{3}\mathrm{NH}_3(g) \quad K_{c1} = 50$

   2. Reverse the reaction Reversing gives:

   $\frac{2}{3}\mathrm{NH}_3(g) \rightleftharpoons \frac{1}{3}\mathrm{N}_2(g) + \mathrm{H}_2(g)$

   For a reversed reaction, $K_c$ becomes the reciprocal:

   $K_{c,\text{reverse}} = \frac{1}{K_{c1}} = \frac{1}{50}$

   3. Multiply the reversed reaction by 3 to clear fractions Multiply every coefficient by 3:

   $2\mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g) + 3\mathrm{H}_2(g)$

   When a reaction is multiplied by a factor $n$, the equilibrium constant is raised to the $n$th power. Here $n = 3$, so: …

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6. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
   Q.The number of Schottky defects present in NaCl per cm$^3$ at room temperature is (A) $10^3$ (B) $10^{23}$ (C) $10^{16}$ (D) $10^6$

   ›Reveal solutionSolution

   Schottky defects are thermally-activated vacancies that occur in ionic crystals. At room temperature, the equilibrium concentration follows Arrhenius behavior, yielding approximately $10^6$ defects per cm³ in NaCl.

   Understanding Schottky Defects

   A Schottky defect occurs when a pair of ions (one cation and one anion) are missing from their lattice sites in an ionic crystal, maintaining electrical neutrality. These defects form spontaneously due to thermal energy and represent an equilibrium between the entropy gain from disorder and the energy cost of creating vacancies.

   The key insight is that defect concentration is governed by thermodynamics, not kinetics. Even though creating a vacancy requires energy, at any finite temperature, some defects will exist because they increase the entropy of the crystal.

   Calculating Equilibrium Defect Concentration

   The number of Schottky defects follows the Arrhenius equation:

   $n = N \exp\left(-\frac{E_s}{2k_BT}\right)$

   where:

   • $n$ = number of Schottky defects per unit volume
   • $N$ = number of lattice sites per unit volume
   • $E_s$ = energy to form one Schottky defect pair
   • $k_B$ = Boltzmann constant
   • $T$ = absolute temperature

   1. Determine the number of lattice sites ($N$)

   For NaCl, there are approximately $2 \times 10^{22}$ ion pairs per cm³ (this comes from the density and molar volume of NaCl).

   2. Identify the formation energy ($E_s$)

   For NaCl, the Schottky defect formation energy is approximately $E_s \approx 2$ eV (this is an experimentally determined value).

   3. Apply room temperature conditions

   At room temperature ($T \approx 300$ K):

   • $k_BT \approx 0.026$ eV
   • Therefore: $\frac{E_s}{2k_BT} = \frac{2}{2 \times 0.026} \approx 38$

   4. Calculate the defect fraction

   $\frac{n}{N} = \exp(-38) \approx 3 \times 10^{-17}$

   5. Find the absolute number of defects …

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7. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.Match the following.

   [!FORMULA] $\begin{array}{ll} \text{Metal Sulfide} & \text{Solubility product} \\ \text{A) PbS} & \text{I) } 4.0 \times 10^{-53} \\ \text{B) HgS} & \text{II) } 8.0 \times 10^{-28} \\ \text{C) MnS} & \text{III) } 1.6 \times 10^{-24} \\ \text{D) ZnS} & \text{IV) } 2.5 \times 10^{-13} \\ \end{array}$

   The correct match is (A) A B C D \quad I II III IV (B) A B C D \quad I IV II III (C) A B C D \quad III IV I II (D) A B C D \quad III IV II I

   ›Reveal solutionSolution

   The solubility product $K_{sp}$ of a metal sulfide depends on the solubility of the sulfide in water, which is linked to the position of the metal in the qualitative analysis scheme. The order of increasing $K_{sp}$ is PbS < HgS < ZnS < MnS, matching A→I, B→II, C→IV, D→III, so the correct option is (B).

   The key concept here is solubility product and its relation to the qualitative analysis of cations. In the classical H₂S scheme, metal sulfides precipitate at different pH values because their $K_{sp}$ values vary enormously. The less soluble the sulfide, the smaller its $K_{sp}$. PbS and HgS are very insoluble (Group II in the H₂S scheme), while MnS is much more soluble (Group IV). ZnS sits in between (Group II but soluble in dilute acid). So the order of increasing $K_{sp}$ (from smallest to largest) is: PbS < HgS < ZnS < MnS.

   Now let’s match the given numbers to this order.

   1. Identify the smallest $K_{sp}$.

      The smallest value among the four is $4.0 \times 10^{-53}$ (I). This must belong to the most insoluble sulfide. Between PbS and HgS, HgS is famously the least soluble — its $K_{sp}$ is astronomically small. So HgS → I is a strong candidate. But let’s check: PbS has $K_{sp} \approx 10^{-28}$ range, so indeed HgS gets the $10^{-53}$ value. So B → I.

   2. Next smallest $K_{sp}$.

      The next smallest is $8.0 \times 10^{-28}$ (II). This fits PbS, which is very insoluble but not as extreme as HgS. So A → II.

   3. The two larger $K_{sp}$ values. …

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8. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
   Q.The ratio of packing density in FCC, BCC, simple cubic and HCP, respectively, is (A) $0.7 : 0.92 : 1.0 : 1.0$ (B) $1.0 : 0.7 : 0.92 : 1.0$ (C) $1.0 : 0.92 : 0.7 : 1.0$ (D) $0.92 : 0.5 : 1.0 : 0.92$

   ›Reveal solutionSolution

   The packing densities are FCC = 0.74, HCP = 0.74, BCC = 0.68, simple cubic = 0.52.

   Their ratio in the order FCC : BCC : simple cubic : HCP is 0.74 : 0.68 : 0.52 : 0.74, which simplifies to 1.0 : 0.92 : 0.7 : 1.0.

   The correct option is (C).

   Concept & Intuition

   Packing density (or atomic packing factor, APF) is the fraction of volume in a crystal structure that is actually occupied by atoms. It depends on two things: how many atoms are in a unit cell, and how efficiently they fill space. For hard spheres of equal size, the densest possible packing is 0.74 — achieved by both FCC and HCP. BCC is slightly less dense (0.68), and simple cubic is the loosest (0.52). The question asks for the ratio of these values in the order FCC, BCC, simple cubic, HCP. So we just compute each APF and compare.

   Step-by-step reasoning

   1. Simple cubic (SC)

      • One atom per unit cell (8 corners × 1/8).
      • Edge length $a = 2r$ (atoms touch along the edge).
      • Volume of atom = $\frac{4}{3}\pi r^3$, cell volume = $a^3 = 8r^3$.
      • APF = $\frac{1 \times \frac{4}{3}\pi r^3}{8r^3} = \frac{\pi}{6} \approx 0.5236$.

   2. Body-centered cubic (BCC)

      • 2 atoms per unit cell (1 center + 8 corners × 1/8).
      • Atoms touch along the body diagonal: $\sqrt{3}a = 4r$ → $a = \frac{4r}{\sqrt{3}}$.
      • Cell volume = $a^3 = \frac{64r^3}{3\sqrt{3}}$.
      • APF = $\frac{2 \times \frac{4}{3}\pi r^3}{64r^3/(3\sqrt{3})} = \frac{\pi\sqrt{3}}{8} \approx 0.6802$.

   3. Face-centered cubic (FCC)

      • 4 atoms per unit cell (6 face centers × 1/2 + 8 corners × 1/8).
      • Atoms touch along the face diagonal: $\sqrt{2}a = 4r$ → $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}\,r$.
      • Cell volume = $a^3 = 16\sqrt{2}\,r^3$.
      • APF = $\frac{4 \times \frac{4}{3}\pi r^3}{16\sqrt{2}\,r^3} = \frac{\pi}{3\sqrt{2}} \approx 0.7405$.

   4. Hexagonal close-packed (HCP)

      • For ideal HCP (c/a ratio = $\sqrt{8/3} \approx 1.633$), the packing density is the same as FCC: 0.7405.
      • Reason: both are close-packed structures; the difference is only in stacking order (ABCABC vs ABAB), not in the fraction of filled space.

   5. Form the ratio …

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9. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.Potassium crystallizes in FCC lattice with unit cell length of $0.5$ nm. The approximate density (in $\mathrm{g\ cm^{-3}}$), if it contains $0.1\%$ Schottky defects is (A) $1.2$ (B) $2.1$ (C) $1.7$ (D) $2.8$

   ›Reveal solutionSolution

   Schottky defects reduce the effective number of atoms in a unit cell, thereby decreasing the crystal's density. For potassium in an FCC lattice with $0.1\%$ Schottky defects, the approximate density is $\boxed{2.1 \text{ g/cm}^3}$.

   Concept and Intuition

   1. Crystal Lattice and Unit Cell: Potassium crystallizes in a Face-Centered Cubic (FCC) lattice. In an FCC unit cell, atoms are located at all eight corners and the center of each of the six faces. The effective number of atoms belonging to one FCC unit cell ($Z$) is calculated as $(8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 1 + 3 = 4$.
   2. Density of a Crystal: The density ($\rho$) of a crystalline solid is determined by the total mass of atoms within a unit cell divided by the volume of the unit cell. The mass of the unit cell is the number of atoms ($Z$) multiplied by the mass of a single atom (Molar mass $M$ divided by Avogadro's number $N_A$). The volume of a cubic unit cell is $a^3$, where $a$ is the unit cell length.
   3. Schottky Defects: A Schottky defect is a type of point defect in a crystal lattice where an atom (or an ion pair in ionic crystals) is missing from its regular lattice site, creating a vacancy.
      • The presence of these missing atoms means that the total mass within a given volume of the crystal is less than that of a perfect crystal.
      • Consequently, Schottky defects decrease the overall density of the crystal.
      • If a crystal has $x\%$ Schottky defects, it implies that $x\%$ of the lattice sites are vacant, and thus, the effective number of atoms contributing to the mass of the unit cell is reduced by $x\%$.

   Step-by-Step Solution

   1. Identify Given Values and Constants:

      • Lattice type: FCC, so the number of atoms per unit cell for a perfect crystal ($Z$) is $4$.
      • Unit cell length ($a$) = $0.5 \text{ nm}$.
      • Percentage of Schottky defects = $0.1\%$.
      • Molar mass of Potassium ($M$) = $39 \text{ g/mol}$ (standard atomic weight).
      • Avogadro's number ($N_A$) = $6.022 \times 10^{23} \text{ mol}^{-1}$.

   2. Convert Unit Cell Length to Centimeters:

      The density is required in $\mathrm{g\ cm^{-3}}$, so we convert the unit cell length from nanometers to centimeters:

      $a = 0.5 \text{ nm} = 0.5 \times 10^{-9} \text{ m} = 0.5 \times 10^{-7} \text{ cm}$.

   3. Calculate the Volume of the Unit Cell:

      For a cubic unit cell, the volume ($V$) is $a^3$:

      $V = (0.5 \times 10^{-7} \text{ cm})^3 = 0.125 \times 10^{-21} \text{ cm}^3$.

   4. Determine the Effective Number of Atoms per Unit Cell ($Z'$) with Schottky Defects:

      A $0.1\%$ Schottky defect means that $0.1\%$ of the lattice sites are vacant. This directly translates to a $0.1\%$ reduction in the effective number of atoms present in the unit cell.

      Fraction of defects = $0.1\% = \frac{0.1}{100} = 0.001$.

      The effective number of atoms per unit cell ($Z'$) is:

      $Z' = Z \times (1 - \text{fraction of defects})$

      $Z' = 4 \times (1 - 0.001) = 4 \times 0.999 = 3.996$. …

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10. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.If the molar concentrations of base and its conjugate acid are same, then pOH of the buffer solution is (A) same as $\mathrm{p}K_b$ of base (B) same as $\mathrm{p}K_a$ of base (C) same as $\mathrm{p}K_a$ of acid (D) same as $\mathrm{p}K_b$ of acid

    ›Reveal solutionSolution

    When the molar concentrations of a weak base and its conjugate acid are equal in a buffer, the Henderson–Hasselbalch equation for bases gives $\text{pOH} = \text{p}K_b$ of the base. The correct option is (A).

    The key here is the Henderson–Hasselbalch equation for a basic buffer. A buffer made from a weak base and its conjugate acid (the salt of that base) resists changes in pH. The equation that governs its pOH is:

    $\text{pOH} = \text{p}K_b + \log\frac{[\text{conjugate acid}]}{[\text{base}]}$

    This is the direct analogue of the acid-buffer equation $\text{pH} = \text{p}K_a + \log\frac{[\text{conjugate base}]}{[\text{acid}]}$. The logic is identical: the ratio of the two species determines how far the pOH is from the base's $\text{p}K_b$.

    Now, the problem states that the molar concentrations of the base and its conjugate acid are the same. That means $[\text{base}] = [\text{conjugate acid}]$, so the ratio $\frac{[\text{conjugate acid}]}{[\text{base}]} = 1$.

    1. Plug this into the Henderson–Hasselbalch equation for bases:

    $\text{pOH} = \text{p}K_b + \log(1)$

    2. Since $\log(1) = 0$, the equation simplifies immediately to:

    $\text{pOH} = \text{p}K_b$

    That’s the entire reasoning — no further calculation needed. The pOH of the buffer equals the $\text{p}K_b$ of the weak base when the two components are at equal concentration. …

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11. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.If the density of a 2 M solution of ethylene glycol in water is 1.11 g/ml, the molality (in 'm') of the solution is approximately (A) 1.92 (B) 1.57 (C) 2.05 (D) 2.15

    ›Reveal solutionSolution

    Take 1 L of solution: mass $=1110$ g, solute $=124$ g, water $=986$ g; molality $=2/0.986 \approx 2.05\ m$.

    Take exactly $1\ \text{L}$ of the $2\ \text{M}$ solution.

    • Moles of ethylene glycol (C$_2$H$_6$O$_2$, $M = 62\ \text{g/mol}$) $= 2\ \text{mol}$.
    • Mass of solute $= 2 \times 62 = 124\ \text{g}$.
    • Mass of solution $= 1000\ \text{mL} \times 1.11\ \text{g/mL} = 1110\ \text{g}$. …

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12. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.The ratio of the highest to the lowest wavelength of Lyman series is (A) $4 : 3$ (B) $9 : 8$ (C) $27 : 5$ (D) $16 : 5$

    ›Reveal solutionSolution

    The highest wavelength in the Lyman series corresponds to the smallest energy transition ($n=2$ to $n=1$), while the lowest wavelength corresponds to the largest energy transition ($n=\infty$ to $n=1$). The ratio of these wavelengths is $\boxed{4:3}$.

    Concept and Intuition

    When an electron in a hydrogen atom (or a hydrogen-like ion) jumps from a higher energy level to a lower one, it emits a photon. The energy of this photon corresponds to the energy difference between the two levels, and this energy determines the photon's wavelength.

    The energy levels in a hydrogen atom are quantized, given by $E_n = -\frac{13.6}{n^2} \text{ eV}$, where $n$ is the principal quantum number ($n=1, 2, 3, \dots$).

    The energy of the emitted photon is $\Delta E = E_{n_i} - E_{n_f}$, where $n_i$ is the initial (higher) energy level and $n_f$ is the final (lower) energy level.

    This energy is also related to the wavelength $\lambda$ by the equation $\Delta E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light.

    From this, we see that $\lambda = \frac{hc}{\Delta E}$. This means:

    • A larger energy difference ($\Delta E$) results in a shorter wavelength ($\lambda$).
    • A smaller energy difference ($\Delta E$) results in a longer wavelength ($\lambda$).

    The Lyman series specifically refers to transitions where electrons fall to the ground state, meaning the final energy level is $n_f = 1$. The initial energy level $n_i$ can be $2, 3, 4, \dots, \infty$.

    To find the highest wavelength ($\lambda_{max}$) in the Lyman series, we need the smallest possible energy difference. This occurs for the transition from $n_i = 2$ to $n_f = 1$.

    To find the lowest wavelength ($\lambda_{min}$) in the Lyman series, we need the largest possible energy difference. This occurs for the transition from $n_i = \infty$ (the ionization limit) to $n_f = 1$.

    We can use the Rydberg formula, which directly relates the wavelength of emitted light to the principal quantum numbers of the initial and final states.

    The Rydberg formula for the wavelength $\lambda$ of spectral lines in a hydrogen atom is given by:

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

    where $R$ is the Rydberg constant, $n_f$ is the principal quantum number of the final energy level, and $n_i$ is the principal quantum number of the initial energy level ($n_i > n_f$).

    Step-by-step Derivation

    1. Identify the Lyman series parameters: For the Lyman series, electrons transition to the ground state. Therefore, the final principal quantum number is $n_f = 1$. The Rydberg formula for the Lyman series becomes:

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

    2. Calculate the highest wavelength ($\lambda_{max}$): The highest wavelength corresponds to the smallest energy transition. For the Lyman series ($n_f=1$), the smallest energy transition occurs when the electron falls from the very next higher level, which is $n_i = 2$. Substitute $n_i = 2$ into the Rydberg formula: $\frac{1}{\lambda_{max}} = R \left( 1 - \frac{1}{2^2} \right)$ …

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