Q.(a) Name the spectral series for a hydrogen atom which lies in the visible region. Find the ratio of the maximum to the minimum wavelengths of this series.

Showing the 12 most recent of 140 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) : If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed. Reason (R) : Electrons behave as both particles and waves.

   ›Reveal solutionSolution

   Electrons exhibit wave-particle duality; their wave nature (de Broglie wavelength) causes diffraction when passing through narrow slits, just like light waves. Both statements are true, and the Reason correctly explains the Assertion.

   Understanding Wave-Particle Duality

   The heart of this question lies in one of quantum mechanics' most profound insights: matter at the atomic scale doesn't fit neatly into our classical categories of "particle" or "wave." Louis de Broglie proposed in 1924 that every moving particle has an associated wavelength, given by

   $\lambda = \frac{h}{p} = \frac{h}{mv}$

   where $h$ is Planck's constant, $p$ is momentum, $m$ is mass, and $v$ is velocity.

   When we accelerate electrons, we increase their momentum. Yet even at high speeds, electrons retain a measurable de Broglie wavelength—typically on the order of angstroms for electrons accelerated through a few hundred volts. This wavelength is comparable to the spacing between atoms in crystals or the width of carefully engineered slits.

   Examining the Assertion

   Assertion (A): If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed.

   This is experimentally verified and true. The classic demonstration is the Davisson-Germer experiment (1927), which showed electron diffraction from crystal lattices. More dramatically, modern versions of the double-slit experiment with electrons—sending them one at a time—build up an interference pattern on a detector screen over time.

   Diffraction occurs when waves encounter obstacles or apertures comparable to their wavelength. The electron beam, despite being composed of particles with mass and charge, produces the characteristic bright and dark fringes we associate with wave phenomena. The central maximum, secondary maxima, and minima all appear exactly as wave theory predicts.

   For single-slit diffraction, minima occur at angles $\theta$ satisfying:

   $a \sin\theta = n\lambda$

   where $a$ is the slit width, $n = 1, 2, 3, \ldots$, and $\lambda$ is the de Broglie wavelength.

   Examining the Reason

   Reason (R): Electrons behave as both particles and waves.

   This is the principle of wave-particle duality, a cornerstone of quantum mechanics. It is unequivocally true.

   Electrons exhibit particle properties: they have definite mass ($9.11 \times 10^{-31}$ kg), charge ($-1.6 \times 10^{-19}$ C), and produce localized impacts on detectors (you can count individual electron arrivals). Simultaneously, they exhibit wave properties: they diffract, interfere, and possess a wavelength and frequency.

   Neither description alone is complete. The electron is a quantum object, and which aspect we observe depends on the experimental setup. When we look for particle behavior (measuring position or momentum), we find particles. When we create conditions for wave behavior (slits, crystals), we observe diffraction and interference.

   Tip

   A useful way to remember this: small wavelength (high momentum) → more particle-like behavior; large wavelength (low momentum) → more wave-like behavior. But both aspects are always present.

   Connecting Assertion and Reason

   Now the critical question: does the Reason explain the Assertion?

   The answer is yes. The diffraction pattern observed when electrons pass through a slit is a direct consequence of their wave nature. Without wave-like behavior, electrons would simply pass through the slit in straight lines, creating a single bright spot on a screen—no fringes, no pattern.

   The wave-particle duality is not just a parallel fact; it is the cause of the diffraction. The de Broglie wavelength associated with the electron's momentum determines the spacing and angular positions of the diffraction maxima and minima. The Reason provides the fundamental physical principle that makes the Assertion true.

   Watch out

   Don't confuse "both are true" with "one explains the other." Many assertion-reason questions have two correct statements that are unrelated. Here, the causal link is explicit: wave nature → diffraction.

   ✓Final answer

   The correct option is (A): Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

2. 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).

   Watch out

   A common mistake is to think that because both statements are true, the reason must automatically explain the assertion. But the reason here is a classical condition that applies to any orbit — it doesn’t pick out only the allowed ones. Quantisation comes from an additional postulate, not from the force balance.

   Tip

   A quick way to test such questions: ask yourself, “If I only knew the reason, would I necessarily conclude the assertion?” Here, knowing only that electrostatic force provides centripetal force, you would not conclude that energy levels are discrete — you’d expect a continuous range. So the reason fails as an explanation.

   4. Match with the given codes:

   • (A): Both true, and reason is correct explanation — No.
   • (B): Both true, but reason is not correct explanation — Yes.
   • (C): Assertion true, reason false — No, reason is true.
   • (D): Both false — No.

   ✓Final answer

   The correct option is (B).

3. 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 $\lambda'$:

   $\frac{hc}{\lambda'} = 13.6 \times \frac{21}{100}$

   Finding the ratio:

   6. Divide the two energy equations:

   $\frac{\lambda'}{\lambda} = \frac{\Delta E_1}{\Delta E_2} = \frac{13.6 \times \frac{24}{25}}{13.6 \times \frac{21}{100}}$

   7. Simplify:

   $\frac{\lambda'}{\lambda} = \frac{\frac{24}{25}}{\frac{21}{100}} = \frac{24}{25} \times \frac{100}{21} = \frac{24 \times 4}{21} = \frac{96}{21} = \frac{32}{7}$

   8. Therefore:

   $\lambda' = \frac{32}{7}\lambda$

   Tip

   The $5 \to 2$ transition releases less energy than $5 \to 1$ (since level 2 is higher than level 1), so the photon must have a longer wavelength. This gives us a quick sanity check: $\lambda' > \lambda$ ✓

   ✓Final answer

   The correct option is (D) $\dfrac{32}{7}\lambda$.

4. CBSE 2026Set 55/3/11 markMCQ
   Q.A proton and an alpha particle have equal momentum. The ratio of their kinetic energies $\left(\dfrac{E_p}{E_\alpha}\right)$ and the ratio of the de Broglie wavelengths associated with them $\left(\dfrac{\lambda_p}{\lambda_\alpha}\right)$ respectively are : (A) $2,\ 1$ (B) $1,\ 2$ (C) $4,\ 1$ (D) $1,\ 4$

   ›Reveal solutionSolution

   For equal momentum, kinetic energy is inversely proportional to mass, and de Broglie wavelength is directly proportional to mass. Since the alpha particle has 4 times the mass of a proton, the ratio of kinetic energies is $4:1$ and the ratio of wavelengths is $1:1$. The correct option is (C).

   The key to this problem lies in two fundamental relationships: the de Broglie wavelength and the connection between kinetic energy and momentum. When two particles have the same momentum, their de Broglie wavelengths become equal — that part is immediate. The kinetic energy, however, depends on mass because $E_k = p^2/2m$, so the lighter particle has more kinetic energy.

   Let’s work through it systematically.

   1. Recall the de Broglie wavelength formula.

      Every moving particle has a wavelength associated with it, given by $\lambda = \frac{h}{p}$, where $h$ is Planck’s constant and $p$ is the linear momentum. This is a direct consequence of wave-particle duality — the more momentum a particle has, the shorter its wavelength.

      Since the problem states that the proton and alpha particle have equal momentum, we can write:

   $p_p = p_\alpha$

   Therefore:

   $\lambda_p = \frac{h}{p_p} \quad \text{and} \quad \lambda_\alpha = \frac{h}{p_\alpha}$

   Because $p_p = p_\alpha$, the two wavelengths are identical:

   $\frac{\lambda_p}{\lambda_\alpha} = 1$

   2. Now find the kinetic energy ratio. Kinetic energy is related to momentum by:

   $E_k = \frac{p^2}{2m}$

   This comes from combining $E_k = \frac12 mv^2$ with $p = mv$. For equal momentum, the kinetic energy is inversely proportional to mass — a heavier particle moving with the same momentum must be slower, so it has less kinetic energy.

   For the proton (mass $m_p$) and alpha particle (mass $m_\alpha$):

   $\frac{E_p}{E_\alpha} = \frac{p^2 / 2m_p}{p^2 / 2m_\alpha} = \frac{m_\alpha}{m_p}$

   3. Know the masses involved. An alpha particle is a helium nucleus — 2 protons and 2 neutrons. Its mass is approximately 4 times the mass of a proton:

   $m_\alpha = 4 m_p$

   This is a standard fact you should remember for such problems.

   Substituting:

   $\frac{E_p}{E_\alpha} = \frac{4 m_p}{m_p} = 4$

   4. Combine the results. We have:

   $\frac{E_p}{E_\alpha} = 4 \quad \text{and} \quad \frac{\lambda_p}{\lambda_\alpha} = 1$

   This matches option (C).

   Watch out

   A common mistake is to forget that kinetic energy depends on mass even when momentum is fixed. Students sometimes assume $E_k \propto p$ and get the ratio wrong. Always write $E_k = p^2/2m$ explicitly.

   Tip

   For any two particles with the same momentum, the de Broglie wavelength ratio is always 1 — it doesn’t matter what the particles are. The kinetic energy ratio is simply the inverse ratio of their masses.

   ✓Final answer

   The correct option is (C), with $\dfrac{E_p}{E_\alpha} = 4$ and $\dfrac{\lambda_p}{\lambda_\alpha} = 1$.

5. CBSE 2026Set V11 markMCQ
   Q.An $\alpha$-particle, a proton, an electron and a neutron are moving with the same velocity. Then the particle having longest de Broglie wavelength is :

   (a) proton

   (b) electron

   (c) neutron

   (d) $\alpha$-particle

   ›Reveal solutionSolution

   (b) electron

   ✓Final answer

   (b) electron

   The de Broglie wavelength is $\lambda = \dfrac{h}{mv}$. For the same velocity $v$, $\lambda \propto \dfrac{1}{m}$, so the particle with the smallest mass has the longest wavelength. Among an $\alpha$-particle, proton, electron and neutron, the electron has by far the smallest mass, so it has the longest de Broglie wavelength.

6. 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) $K > 0,\ U < 0,\ E < 0$.

   ✓Final answer

   Option (b) $K > 0,\ U < 0,\ E < 0$.

   For an electron bound in a hydrogen atom the kinetic energy is always positive, the electrostatic potential energy (electron–nucleus attraction, taken zero at infinity) is negative, and the total energy $E = K + U = -K$ is negative because the electron is bound. In fact $U = -2K$ and $E = -K$.

7. 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.

   For the hydrogen atom (Bohr model), the total energy of the electron in the nth stationary state is $E_n=-\dfrac{13.6}{n^2}\text{ eV}$ (negative, since the electron is bound).

   ✓Final answer

   (iii) $-\dfrac{13.6}{n^2}$ eV

8. CBSE 2026Set A1 markMCQ
   Q.The ratio of de Broglie wavelength associated with two electrons accelerated through 49 V and 64 V is (A) 49/64 (B) 64/49 (C) 7/8 (D) 8/7

   ›Reveal solutionSolution

   de Broglie wavelength of an accelerated electron λ ∝ 1/√V; ratio = √(64/49) = 8/7.

   For an electron accelerated through potential difference V, its kinetic energy is $eV = \frac{p^2}{2m}$, so $p = \sqrt{2meV}$ and the de Broglie wavelength is

   $\lambda = \frac{h}{\sqrt{2meV}} \propto \frac{1}{\sqrt{V}}.$

   Hence

   $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{64}{49}} = \frac{8}{7}.$

   (The electron through the larger voltage, 64 V, has the shorter wavelength.)

   ✓Final answer

   (D) 8/7.

9. 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 continuous background. Hence the solar spectrum is best described as an absorption spectrum of black (dark) lines.

   ✓Final answer

   (C) spectrum of black lines.

10. 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

    $K = \tfrac{1}{2} m v^2 = \frac{k e^2}{2r}.$

    The potential energy (taking zero at infinity) is

    $U = -\frac{k e^2}{r}.$

    Comparing, $K = -\dfrac{U}{2}$. (The total energy is $E = K + U = -\dfrac{k e^2}{2r} = -K$.)

    ✓Final answer

    (C) K = -U/2.

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.If alpha particle, proton and electron move with the same momentum, then their respective de Broglie wavelengths λα, λp, λe are related as

    (a) λα > λp > λe

    (b) λα < λp < λe

    (c) λα = λp = λe

    (d) None of the above

    ›Reveal solutionSolution

    Equal momentum means equal de Broglie wavelength, regardless of the particles' different masses.

    The de Broglie wavelength of a particle is

    $\lambda = \dfrac{h}{p}$

    where $h$ is Planck's constant and $p$ is the particle's momentum. This formula depends only on momentum $p$, not on the particle's mass, charge, or identity. Since the alpha particle, proton and electron are all stated to have the same momentum, they must all have the same de Broglie wavelength, even though their masses (and hence their speeds and kinetic energies, for the same $p$) are very different.

    ✓Final answer

    (c) $\lambda_\alpha = \lambda_p = \lambda_e$.

12. CBSE 2026Set ANNUAL1 mark
    Q.A body of mass 0.10 kg is moving with a speed of 10 m/s. The de-Broglie wavelength of the wave associated with it will be ______ m.

    ›Reveal solutionSolution

    The de Broglie wavelength of a moving body is lambda = h/(mv); plugging in the mass and speed gives the value directly.

    De Broglie's relation: lambda = h / (m v), where h = 6.63 x 10^-34 J.s (Planck's constant), m = 0.10 kg, v = 10 m/s.

    mv = 0.10 x 10 = 1 kg.m/s

    lambda = 6.63 x 10^-34 / 1 = 6.63 x 10^-34 m

    ✓Final answer

    lambda = 6.63 x 10^-34 m.