Q.Assertion (A): D(+)-Glucose is dextrorotatory in nature.
Reason (R): 'D' represents its dextrorotatory nature.

Showing the 12 most recent of 53 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Law of conservation of mass was regarded as another basic conservation law of nature until the advent of (A) Bose-Einstein statistics (B) Theory of relativity (C) Chandrasekhar limit (D) Uncertainty principle

   ›Reveal solutionSolution

   The law of conservation of mass was overturned as a fundamental principle by Einstein’s theory of relativity, which showed mass and energy are interchangeable. The correct option is (B).

   The key idea here is that before the 20th century, scientists believed mass could neither be created nor destroyed — it was a bedrock of chemistry and physics. But the theory of relativity changed that by revealing that mass is a form of energy, so mass can be “lost” when converted into energy (as in nuclear reactions). The other options deal with quantum statistics, stellar limits, or quantum uncertainty — none directly challenge mass conservation.

   Let’s walk through each option:

   1. Bose-Einstein statistics — This describes how identical particles (bosons) behave at low temperatures, leading to phenomena like Bose-Einstein condensates. It doesn’t alter the conservation of mass; it’s about quantum statistical behavior, not mass-energy equivalence.

   2. Theory of relativity — Einstein’s special relativity (1905) introduced $E = mc^2$, showing that mass and energy are two sides of the same coin. In nuclear reactions, a tiny amount of mass disappears and reappears as a huge amount of energy. This directly breaks the classical law of conservation of mass, replacing it with the more general conservation of mass-energy.

   3. Chandrasekhar limit — This is the maximum mass (about 1.4 solar masses) a white dwarf star can have before collapsing into a neutron star. It’s a consequence of quantum mechanics and gravity, not a challenge to mass conservation itself. …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The ratio of wavenumber of first line of Balmer series and wavenumber of second line of Lyman series of hydrogen atom is (A) $5:32$ (B) $32:5$ (C) $22:7$ (D) $7:22$

   ›Reveal solutionSolution

   The wavenumber ratio is found by applying the Rydberg formula to the first Balmer line (transition from n=3 to n=2) and the second Lyman line (transition from n=4 to n=1). The ratio simplifies to 5:32, so the correct option is (A).

   The key concept here is the Rydberg formula for wavenumber of spectral lines in hydrogen. The wavenumber $\tilde{\nu}$ (inverse wavelength) for a transition from a higher energy level $n_2$ to a lower level $n_1$ is given by:

   $\tilde{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

   where $R_H$ is the Rydberg constant. The Balmer series corresponds to transitions ending at $n_1 = 2$, and the Lyman series to transitions ending at $n_1 = 1$. The "first line" of a series means the transition with the smallest energy difference (longest wavelength), and the "second line" means the next smallest.

   Let’s work through the calculation step by step.

   1. Identify the transitions.
      • First line of Balmer series: This is the transition from $n_2 = 3$ to $n_1 = 2$.
      • Second line of Lyman series: This is the transition from $n_2 = 4$ to $n_1 = 1$. (The first Lyman line is from $n=2$ to $n=1$; the second is from $n=3$ to $n=1$; wait — careful: the second line of Lyman is actually from $n=3$ to $n=1$? Let’s double-check: Lyman series: $n_1=1$. The first line: $n_2=2$. The second line: $n_2=3$. The third line: $n_2=4$. So the second line of Lyman is from $n=3$ to $n=1$, not $n=4$. This is a common pitfall — see the warning below.)

   Watch out

   A classic mistake is to think the "second line" means the second possible upper level, which is $n=3$ for Lyman, not $n=4$. The first line uses $n=2$, the second uses $n=3$, the third uses $n=4$, etc. So for the second Lyman line, $n_2 = 3$.

   2. Write the wavenumber for each line.
      • For the first Balmer line ($n_1=2, n_2=3$): $$\tilde{\nu}_B = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9-4}{36} \right) = R_H \cdot \frac{5}{36} …$$

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the activities of a radioactive substance at times $t = 0$ and $t = 3T$ are $A$ and $B$ respectively, then the activity of the substance at a time $t = 9T$ is (A) $\dfrac{A^3}{B^2}$ (B) $\dfrac{A^2}{B}$ (C) $\dfrac{B^2}{A}$ (D) $\dfrac{B^3}{A^2}$

   ›Reveal solutionSolution

   Radioactive decay follows an exponential law: activity $R(t)=R_0 e^{-\lambda t}$. Using the given values at $t=0$ and $t=3T$, we find $e^{-\lambda T} = (B/A)^{1/3}$. Then activity at $t=9T$ becomes $A \cdot (B/A)^3 = B^3/A^2$, which matches option (D).

   ---

   The key idea is that radioactive decay is exponential: the activity decreases by a constant factor over equal time intervals. So if we know the factor over $3T$, we can raise it to the appropriate power to get the factor over $9T$.

   1. Write the decay law

      Activity at time $t$ is $R(t) = R_0 e^{-\lambda t}$, where $R_0$ is the initial activity and $\lambda$ the decay constant.

      At $t=0$: $R(0) = R_0 = A$.

      At $t=3T$: $R(3T) = A e^{-\lambda \cdot 3T} = B$.

   2. Find the decay factor over $3T$

      From $A e^{-3\lambda T} = B$, we get

   $e^{-3\lambda T} = \frac{B}{A}.$

   Taking the cube root:

   $e^{-\lambda T} = \left(\frac{B}{A}\right)^{1/3}.$

   This is the factor by which activity drops every $T$ seconds.

   3. Activity at $t = 9T$ $9T$ is three intervals of $3T$, or nine intervals of $T$. Using the factor per $T$: $R(9T) = A \cdot \left(e^{-\lambda T}\right)^9 = A \cdot \left[\left(\frac{B}{A}\right)^{1/3}\right]^9.$ …

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4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The velocity of the photoelectron (in ms$^{-1}$) emitted when a smooth surface of a metal is made to strike with a photon of wavelength $4 \times 10^{-7}$ m is ($W_0$ of metal = 2.13 eV; 1 eV = $1.6 \times 10^{-19}$ J ; $h = 6.6 \times 10^{-34}$ Js ; $m_e = 9.1 \times 10^{-31}$ kg) (A) $\dfrac{1}{\sqrt{5}} \times 10^6$ (B) $\dfrac{1}{\sqrt{3}} \times 10^5$ (C) $\dfrac{1}{\sqrt{3}} \times 10^6$ (D) $\dfrac{1}{\sqrt{7}} \times 10^5$

   ›Reveal solutionSolution

   The photoelectron’s velocity is found from Einstein’s photoelectric equation: kinetic energy = photon energy minus work function. After converting units and solving, the speed is $\frac{1}{\sqrt{3}} \times 10^6$ m/s, which matches option (C).

   The core idea is Einstein’s photoelectric effect: a photon gives all its energy to an electron. Some of that energy is used to overcome the work function $W_0$ (the minimum energy to free the electron), and the rest becomes the electron’s kinetic energy. Since the electron is non‑relativistic here, we use $K = \frac12 m v^2$. The question gives wavelength, so we first find photon energy from $E = hc/\lambda$.

   1. Find the photon energy in joules. Photon wavelength $\lambda = 4 \times 10^{-7}$ m.

   $E_{\text{photon}} = \frac{hc}{\lambda}$

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

   $E_{\text{photon}} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}} = \frac{19.8 \times 10^{-26}}{4 \times 10^{-7}} = 4.95 \times 10^{-19} \text{ J}$

   2. Convert the work function to joules. Given $W_0 = 2.13$ eV and $1$ eV = $1.6 \times 10^{-19}$ J:

   $W_0 = 2.13 \times 1.6 \times 10^{-19} = 3.408 \times 10^{-19} \text{ J}$

   3. Apply Einstein’s photoelectric equation. Kinetic energy of the emitted electron:

   $K = E_{\text{photon}} - W_0 = (4.95 - 3.408) \times 10^{-19} = 1.542 \times 10^{-19} \text{ J}$

   4. Relate kinetic energy to velocity.

   $\frac12 m_e v^2 = K \quad\Rightarrow\quad v = \sqrt{\frac{2K}{m_e}}$

   With $m_e = 9.1 \times 10^{-31}$ kg:

   $v = \sqrt{\frac{2 \times 1.542 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{\frac{3.084 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{0.3389 \times 10^{12}}$

   $v = \sqrt{3.389 \times 10^{11}} \approx \sqrt{3.389} \times 10^{5.5}$

   More neatly: $3.389 \times 10^{11} = \frac{10^{12}}{2.95}$, but let’s compute exactly. …

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5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The kinetic energy of photo electron emitted from the surface of a metal is $7.2 \times 10^{-20}$ J, when the metal is made to strike with light having wavelength $x$ nm. What is the value of $x$? (Work function of metal = $4.5$ eV; $h = 6.6 \times 10^{-34}$ Js; $c = 3 \times 10^8$ ms$^{-1}$; $1$ eV $= 1.6 \times 10^{-19}$ J) (A) 125 (B) 250 (C) 175 (D) 150

   ›Reveal solutionSolution

   We use Einstein's photoelectric equation, $KE = E - \phi$, to find the incident photon energy, ensuring all terms are in Joules. Then, we use $E = hc/\lambda$ to calculate the wavelength. The value of $x$ is 250 nm.

   The photoelectric effect is a fundamental phenomenon in quantum physics where electrons are emitted from a material when light shines on it. This effect demonstrates the particle nature of light, where light energy is carried in discrete packets called photons.

   When a photon strikes the surface of a metal, it transfers its energy to an electron. For an electron to be ejected, the photon's energy ($E$) must be greater than or equal to the work function ($\phi$) of the metal. The work function is the minimum energy required to remove an electron from the surface of the metal. Any energy supplied by the photon beyond this work function is converted into the kinetic energy ($KE$) of the emitted electron.

   This energy conservation principle is described by Einstein's photoelectric equation:

   $KE = E - \phi$

   Where:

   • $KE$ is the maximum kinetic energy of the emitted photoelectron.
   • $E$ is the energy of the incident photon.
   • $\phi$ is the work function of the metal.

   The energy of an incident photon is also related to its wavelength ($\lambda$) by the formula:

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

   Where:

   • $h$ is Planck's constant.
   • $c$ is the speed of light in vacuum.

   The crucial step in solving such problems is to ensure all energy terms are expressed in consistent units, typically Joules (J), before performing calculations. The work function is often given in electron volts (eV), which must be converted to Joules.

   Step-by-Step Solution:

   1. Identify given values and the target:

      We are provided with the following information:

      • Kinetic energy of the photoelectrons, $KE = 7.2 \times 10^{-20}$ J.
      • Work function of the metal, $\phi = 4.5$ eV.
      • Planck's constant, $h = 6.6 \times 10^{-34}$ Js.
      • Speed of light, $c = 3 \times 10^8$ ms$^{-1}$.
      • Conversion factor, $1$ eV $= 1.6 \times 10^{-19}$ J. We need to find the wavelength of the incident light, $x$ nm. This means we will calculate the wavelength $\lambda$ in meters and then convert it to nanometers.

   2. Convert the work function to Joules:

      Since the kinetic energy is given in Joules, we must convert the work function from electron volts to Joules to maintain unit consistency in the photoelectric equation.

      $\phi = 4.5 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV})$

      $\phi = 7.2 \times 10^{-19}$ J

   3. Calculate the energy of the incident photon ($E$):

      Using Einstein's photoelectric equation, $KE = E - \phi$, we can rearrange it to solve for the incident photon energy $E$:

      $E = KE + \phi$

      Substitute the given kinetic energy and the converted work function:

      $E = (7.2 \times 10^{-20} \text{ J}) + (7.2 \times 10^{-19} \text{ J})$ …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The work function of three metals A, B and C is respectively 2.25, 2.42 and 3.6 eV. All the three metals were irradiated with light of wavelength 330 nm. The kinetic energy of photoelectrons emitted from A, B and C is respectively $E_A, E_B$ and $E_C$. The correct relationship of $E_A, E_B$ and $E_C$ is $(h = 6.6 \times 10^{-34} \ \text{Js}; \ c = 3 \times 10^8 \ \text{ms}^{-1})$ (A) $E_A > E_B > E_C$ (B) $E_A = E_B = E_C$ (C) $E_A < E_B < E_C$ (D) $E_C > E_A > E_B$

   ›Reveal solutionSolution

   The photoelectric effect tells us that kinetic energy equals photon energy minus work function.

   For a fixed wavelength (330 nm), the photon energy is constant, so the metal with the smallest work function gives the largest kinetic energy.

   Since work functions are A = 2.25 eV, B = 2.42 eV, C = 3.6 eV, we get $E_A > E_B > E_C$.

   The correct option is (A).

   ---

   Concept & Intuition

   The photoelectric effect is beautifully simple: a photon gives all its energy to an electron. The electron uses some of that energy to escape the metal (the work function $\phi$), and the rest becomes kinetic energy.

   So for each metal:

   $K_{\text{max}} = hf - \phi$

   Here, the light’s wavelength is fixed (330 nm), so every photon carries the same energy. That means the metal with the smallest work function will give the electron the most leftover kinetic energy. It’s like three toll booths with different tolls — if you have the same amount of cash, the one with the smallest toll leaves you with the most change.

   ---

   Step‑by‑step reasoning

   1. Find the photon energy The energy of a single photon is

   $E_{\text{photon}} = \frac{hc}{\lambda}$

   Given $h = 6.6 \times 10^{-34}\ \text{Js}$, $c = 3 \times 10^8\ \text{m/s}$, and $\lambda = 330\ \text{nm} = 330 \times 10^{-9}\ \text{m}$:

   $E_{\text{photon}} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{330 \times 10^{-9}} = \frac{1.98 \times 10^{-25}}{3.3 \times 10^{-7}} = 6.0 \times 10^{-19}\ \text{J}$

   Convert to electronvolts (1 eV = $1.6 \times 10^{-19}\ \text{J}$):

   $E_{\text{photon}} = \frac{6.0 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.75\ \text{eV}$

   2. Apply the photoelectric equation for each metal

   $E_A = 3.75 - 2.25 = 1.50\ \text{eV}$

   $$$$

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7. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.The scientific principle useful for the working of a fusion test reactor is (A) Trapping and cooling of atoms by laser beams and magnetic fields (B) Controlled nuclear fission (C) Magnetic confinement of plasma (D) Wave nature of electrons

   ›Reveal solutionSolution

   Fusion test reactors aim to harness nuclear fusion, which requires extremely high temperatures to create plasma; this plasma is then confined and controlled using powerful magnetic fields. The correct option is (C).

   Concept and Intuition

   Nuclear fusion is the process by which two light atomic nuclei combine to form a heavier nucleus, releasing a tremendous amount of energy. This is the same process that powers the Sun and other stars. On Earth, achieving controlled nuclear fusion is incredibly challenging because atomic nuclei are positively charged and repel each other strongly (due to electrostatic repulsion). To overcome this repulsion and allow them to fuse, the nuclei must be brought very close together with immense kinetic energy.

   This requires extremely high temperatures, typically tens to hundreds of millions of degrees Celsius. At such extreme temperatures, atoms ionize, meaning their electrons are stripped away, forming a superheated, electrically charged gas called plasma. This plasma is the "fuel" for fusion.

   The core challenge for a fusion reactor is twofold:

   1. Heating: Reaching and maintaining these extreme temperatures.
   2. Confinement: Containing this superheated plasma for long enough and at sufficient density for fusion reactions to occur efficiently. No physical material can withstand temperatures of millions of degrees, so traditional containers are impossible. This is where the scientific principle of confinement becomes critical.

   Step-by-Step Analysis

   Let's examine each option in the context of a fusion test reactor:

   1. Trapping and cooling of atoms by laser beams and magnetic fields

      • This technique is primarily used in fields like quantum physics to study ultra-cold atoms, create Bose-Einstein condensates, or develop highly precise atomic clocks. It involves slowing down and cooling atoms to temperatures very close to absolute zero.
      • Why it's incorrect for fusion: Fusion requires heating matter to millions of degrees Celsius, not cooling it. The goal is to give nuclei enough energy to overcome their electrostatic repulsion, which is the opposite of cooling.

   2. Controlled nuclear fission

      • Nuclear fission is the process where a heavy atomic nucleus (like uranium or plutonium) is split into two or more smaller nuclei, releasing energy. This is the principle behind current commercial nuclear power plants and atomic bombs.
      • Why it's incorrect for fusion: While both fission and fusion are nuclear reactions that release energy, they are fundamentally different processes. Fusion involves combining light nuclei, whereas fission involves splitting heavy nuclei. A fusion reactor aims to achieve fusion, not fission.

   3. Magnetic confinement of plasma

      • As discussed, fusion requires plasma at extremely high temperatures. This plasma is too hot to be contained by any material walls, as it would instantly melt or vaporize them and cool down rapidly, stopping the fusion reactions. …

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8. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The concept that provided a convincing evidence of the atomic picture of matter is (A) Photo electric effect (B) Brownian motion theory (C) Superconductivity (D) Wave nature of electrons

   ›Reveal solutionSolution

   The concept that gave the most direct and convincing evidence for the atomic (discrete) picture of matter was the Brownian motion theory, which explained the jittery motion of visible particles as the result of invisible molecular collisions.

   The question asks for the concept that provided convincing evidence of the atomic picture of matter — that is, proof that matter is made of discrete, indivisible particles (atoms or molecules), not a continuous substance.

   Let’s look at each option carefully.

   1. Photo electric effect — This showed that light behaves as discrete packets of energy (photons), and that electrons are emitted from a metal surface only when light of a certain minimum frequency strikes it. While this was revolutionary for quantum theory and the particle nature of light, it did not directly prove that matter itself is made of atoms. It proved something about light, not about the structure of matter.

   2. Brownian motion theory — In 1827, Robert Brown observed that pollen grains suspended in water moved in a constant, random, zigzag path. Einstein (1905) and later Perrin (1908) showed theoretically and experimentally that this motion is caused by the continuous, random bombardment of the pollen grains by invisible water molecules. The quantitative agreement between theory and experiment — especially Perrin’s measurement of Avogadro’s number — was the first direct, convincing proof that molecules (and hence atoms) really exist. This is the correct answer.

   3. Superconductivity — This is a macroscopic quantum phenomenon where certain materials conduct electricity with zero resistance below a critical temperature. It reveals important quantum properties of matter but does not serve as evidence for the atomic (discrete) nature of matter itself. It assumes atoms exist; it doesn’t prove they do. …

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9. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The RMS velocity of dihydrogen is $\sqrt{7}$ times more than that of dinitrogen. If $T_{H_2}$ and $T_{N_2}$ are the temperatures of dihydrogen and dinitrogen, then the correct relationship between them is (A) $T_{H_2} = T_{N_2}$ (B) $T_{H_2} > T_{N_2}$ (C) $T_{H_2} = \sqrt{7} T_{N_2}$ (D) $T_{H_2} = \dfrac{T_{N_2}}{2}$

   ›Reveal solutionSolution

   The RMS velocity formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ shows that for a given gas, velocity depends on temperature and molar mass. Given $v_{H_2} = \sqrt{7}\, v_{N_2}$, we find $T_{H_2} = \frac{1}{2} T_{N_2}$, so the correct option is (D).

   The key concept here is the root-mean-square (RMS) velocity of gas molecules, given by

   $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

   where $R$ is the gas constant, $T$ is the absolute temperature, and $M$ is the molar mass.

   The problem tells us that the RMS velocity of dihydrogen ($H_2$) is $\sqrt{7}$ times that of dinitrogen ($N_2$). That means

   $v_{H_2} = \sqrt{7} \, v_{N_2}$

   We need to relate their temperatures. The trick: the molar masses are different — $M_{H_2} = 2\ \text{g/mol}$ and $M_{N_2} = 28\ \text{g/mol}$. So the ratio of velocities depends on both $T$ and $M$. Let’s work it out step by step.

   1. Write the RMS velocity expressions for each gas.

   $v_{H_2} = \sqrt{\frac{3R T_{H_2}}{M_{H_2}}}, \quad v_{N_2} = \sqrt{\frac{3R T_{N_2}}{M_{N_2}}}$

   2. Use the given relation $v_{H_2} = \sqrt{7}\, v_{N_2}$. Substitute the formulas:

   $\sqrt{\frac{3R T_{H_2}}{M_{H_2}}} = \sqrt{7} \cdot \sqrt{\frac{3R T_{N_2}}{M_{N_2}}}$

   3. Square both sides to remove square roots.

   $\frac{3R T_{H_2}}{M_{H_2}} = 7 \cdot \frac{3R T_{N_2}}{M_{N_2}}$

   Cancel $3R$ from both sides:

   $\frac{T_{H_2}}{M_{H_2}} = 7 \cdot \frac{T_{N_2}}{M_{N_2}}$

   4. Plug in the molar masses. $M_{H_2} = 2$, $M_{N_2} = 28$:

   $\frac{T_{H_2}}{2} = 7 \cdot \frac{T_{N_2}}{28}$

   5. Simplify the right-hand side. $\frac{T_{H_2}}{2} = \frac{7}{28} T_{N_2} = \frac{1}{4} T_{N_2}$ …

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10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.The potential energy of an electron in an orbit of hydrogen atom is $-6.8$ eV. The de Broglie wavelength of the electron in this orbit is ($r_0$ is Bohr radius) (A) $2\pi r_0$ (B) $4\pi r_0$ (C) $\pi r_0$ (D) $3\pi r_0$

    ›Reveal solutionSolution

    The key idea is that the de Broglie wavelength of an electron in a Bohr orbit equals the circumference of that orbit divided by the principal quantum number $n$. Given the potential energy $-6.8$ eV, we find $n=2$, so the wavelength is $4\pi r_0$, which corresponds to option (B).

    ---

    Concept and Intuition

    In the Bohr model of the hydrogen atom, an electron moves in circular orbits around the nucleus. A central condition is that the angular momentum is quantized:

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

    where $n = 1, 2, 3, \dots$ is the principal quantum number.

    The de Broglie wavelength of the electron is $\lambda = h / (mv)$. Combining these, we get:

    $\lambda = \frac{h}{mv} = \frac{2\pi r}{n}$

    So the wavelength is simply the circumference of the orbit divided by $n$. This means that if we know the orbit's radius (in terms of the Bohr radius $r_0$) and the quantum number $n$, we can find $\lambda$.

    The problem gives the potential energy $U = -6.8$ eV. In the Bohr model, the total energy of an electron in the $n$-th orbit is:

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

    and the potential energy is twice the total energy (by the virial theorem for a Coulomb force):

    $U = 2E_n = -\frac{27.2}{n^2} \text{ eV}$

    Thus, from $U = -6.8$ eV, we can solve for $n$, then find the radius $r_n = n^2 r_0$, and finally the de Broglie wavelength.

    ---

    Step-by-Step Solution

    1. Find the principal quantum number $n$ from the potential energy. The potential energy in the $n$-th Bohr orbit is:

    $U = -\frac{27.2}{n^2} \text{ eV}$

    Set this equal to $-6.8$ eV:

    $-\frac{27.2}{n^2} = -6.8$

    Cancel the negatives and solve:

    $\frac{27.2}{n^2} = 6.8 \quad \Rightarrow \quad n^2 = \frac{27.2}{6.8} = 4$

    So $n = 2$.

    2. Determine the radius of the orbit. The radius of the $n$-th Bohr orbit is:

    $r_n = n^2 r_0$

    For $n=2$:

    $$$$

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.In the atomic spectrum of hydrogen, the wavelengths of the spectral lines corresponding to electronic transitions

    (i) $n = 4$ to $n = 2$ and

    (ii) $n = 3$ to $n = 1$ are $\lambda_1$ and $\lambda_2$ $\text{\AA}$ respectively. The value of $(\lambda_1 - \lambda_2)$ (in cm) is ($R_H$ = Rydberg constant) (A) $\dfrac{1}{R_H} \left[ \dfrac{24}{101} \right]$ (B) $R_H \left[ \dfrac{24}{101} \right]$ (C) $\dfrac{1}{R_H} \left[ \dfrac{101}{24} \right]$ (D) $R_H \left[ \dfrac{101}{24} \right]$

    ›Reveal solutionSolution

    The key idea is to use the Rydberg formula for hydrogen to express each wavelength in terms of $R_H$, then subtract them and simplify. The result is $\lambda_1 - \lambda_2 = \frac{1}{R_H} \cdot \frac{101}{24}$, so the correct option is (C).

    The problem asks for the difference in wavelengths (in cm) between two spectral lines of hydrogen, given in terms of the Rydberg constant $R_H$. The Rydberg formula directly gives the wavenumber (inverse wavelength) for any transition, so we can compute each wavelength, subtract, and simplify. The trick is to handle the units carefully — the answer choices are expressed as fractions involving $R_H$, so we keep everything symbolic.

    Why this approach works:

    The Rydberg formula is the fundamental relation for hydrogen spectral lines. It expresses the wavenumber $\tilde{\nu} = 1/\lambda$ as $R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. By writing each $\lambda$ as the reciprocal of this, we can directly compute $\lambda_1 - \lambda_2$ algebraically. The answer choices are all in the form of either $R_H$ times a fraction or $1/R_H$ times a fraction, so our result will match one of these.

    ---

    1. Write the Rydberg formula for each transition. For hydrogen, the wavenumber is:

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

    where $R_H$ is the Rydberg constant (in cm$^{-1}$ if $\lambda$ is in cm).

    • Transition (i): $n_i = 4$, $n_f = 2$

    $\frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{4}{16} - \frac{1}{16} \right) = R_H \cdot \frac{3}{16}$

     So:
    

    $\lambda_1 = \frac{1}{R_H} \cdot \frac{16}{3}$

    • Transition (ii): $n_i = 3$, $n_f = 1$

    $\frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \cdot \frac{8}{9}$

     So:
    

    $\lambda_2 = \frac{1}{R_H} \cdot \frac{9}{8}$

    2. Compute the difference $\lambda_1 - \lambda_2$. Both are expressed as multiples of $1/R_H$, so:

    $\lambda_1 - \lambda_2 = \frac{1}{R_H} \left( \frac{16}{3} - \frac{9}{8} \right)$

    Find a common denominator (24):

    $$$$

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.The potential energy of an electron in an orbit of hydrogen atom is $-6.8$ eV. The de Broglie wavelength of the electron in this orbit is ($r_0$ is Bohr radius) (A) $3\pi r_0$ (B) $\pi r_0$ (C) $2\pi r_0$ (D) $4\pi r_0$

    ›Reveal solutionSolution

    The de Broglie wavelength of an electron in a Bohr orbit equals $\dfrac{2\pi r_n}{n}$. The given potential energy $-6.8$ eV identifies $n=2$, and the radius $r_2=4r_0$, giving $\lambda = 4\pi r_0$ — option (D).

    ---

    Concept and Intuition

    The de Broglie wavelength of a particle is $\lambda = h/p$, where $p$ is its momentum. In the Bohr model of hydrogen, a stable orbit must support a standing wave around its circumference — the circumference must be an integer number of wavelengths: $2\pi r_n = n\lambda$, so $\lambda = \dfrac{2\pi r_n}{n}$.

    Given the potential energy, we first find which orbit ($n$) the electron occupies, then find that orbit's radius $r_n = n^2 r_0$, and finally compute $\lambda$.

    ---

    Step-by-step solution

    1. Relate potential energy to the orbit number The total energy of the $n$th Bohr orbit is $E_n = -\dfrac{13.6}{n^2}$ eV. By the virial theorem, the potential energy is twice the total energy:

    $U = 2E_n = -\frac{27.2}{n^2} \text{ eV}.$

    Given $U = -6.8$ eV,

    $-\frac{27.2}{n^2} = -6.8 \quad\Rightarrow\quad n^2 = \frac{27.2}{6.8} = 4 \quad\Rightarrow\quad n = 2.$

    2. Find the radius of the $n=2$ orbit The Bohr radius $r_0$ is the radius of the $n=1$ orbit, and $r_n = n^2 r_0$. So

    $r_2 = 4r_0.$

    3. Apply the standing-wave (Bohr quantization) condition …

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