Q.Using Bohr's atomic model, derive the expression for the radius of $n^{th}$ orbit of the revolving electron in a hydrogen atom.

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1. CBSE 2026Set 55/2/11 markMCQ
   Q.The maximum kinetic energy of the electrons emitted from a photosensitive surface depends on (A) work function of the surface $\phi_0$ only. (B) frequency of the incident radiation $\nu$ only. (C) intensity of the incident radiation $I$ only. (D) Both $\phi_0$ and $\nu$.

   ›Reveal solutionSolution

   The maximum kinetic energy of photoelectrons is determined by Einstein's photoelectric equation: it depends on both the photon energy (set by frequency $\nu$) and the work function $\phi_0$ of the material. The answer is (D).

   Why the photoelectric effect reveals energy quantization

   When light strikes a metal surface, electrons can be ejected—but not in the way classical wave theory predicted. Einstein's revolutionary insight was that light arrives in discrete packets (photons), each carrying energy $E = h\nu$. An electron absorbs one photon entirely; if that energy exceeds the minimum needed to escape the metal (the work function $\phi_0$), the electron breaks free, and any leftover energy becomes kinetic energy.

   This is fundamentally an energy-balance problem. The photon delivers a fixed amount of energy $h\nu$. The electron must "pay" $\phi_0$ to escape. What remains is the maximum kinetic energy:

   $K_{\text{max}} = h\nu - \phi_0$

   Notice what this equation tells us: $K_{\text{max}}$ increases linearly with frequency $\nu$ and decreases with larger work function $\phi_0$. Both parameters matter.

   Step-by-step reasoning

   1. The photon energy is $h\nu$.

      A single photon of frequency $\nu$ carries energy proportional to that frequency. Higher frequency means more energetic photons (ultraviolet photons pack more punch than red photons).

   2. The work function $\phi_0$ is the escape barrier.

      Different materials bind their electrons with different strengths. Cesium has a low work function (~2 eV), so even visible light can eject electrons. Platinum has a high work function (~6 eV), requiring ultraviolet light. This material property directly subtracts from the available kinetic energy.

   3. Energy conservation gives $K_{\text{max}} = h\nu - \phi_0$.

      The electron that escapes with maximum kinetic energy is one that was initially at the Fermi level (loosest bound) and lost no energy to collisions on the way out. All the "profit" after paying $\phi_0$ goes into kinetic energy.

   4. Intensity does not affect $K_{\text{max}}$.

      Intensity measures the number of photons arriving per unit time, not the energy of each photon. Brighter light ejects more electrons (higher photocurrent), but each electron still gets energy from just one photon. Doubling intensity doubles the electron count, not their individual speeds. …

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2. CBSE 2026Set 55/2/11 markMCQ
   Q.In Bohr model of hydrogen atom, for large values of $n$, the distance between the consecutive orbits is proportional to (A) $\sqrt{n}$ (B) $n$ (C) $n^2$ (D) $n^3$

   ›Reveal solutionSolution

   In the Bohr model, the orbital radius scales as $r_n \propto n^2$. For large $n$, the gap between consecutive orbits $\Delta r = r_{n+1} - r_n$ behaves like $2n + 1$, which is proportional to $n$ — so the answer is (B).

   The Bohr model gives a beautifully simple picture of the hydrogen atom: electrons orbit the nucleus in fixed circular paths, with quantised angular momentum. The radius of the $n$-th orbit is

   $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \quad \text{or, more compactly,} \quad r_n = a_0 n^2,$

   where $a_0 \approx 0.529 \, \text{Å}$ is the Bohr radius. So the radius grows as $n^2$.

   The question asks about the distance between consecutive orbits — that is, the difference $r_{n+1} - r_n$ — for large $n$. Many students instinctively think this difference is constant, or that it grows like $n^2$ because the radii themselves do. But a difference between two quadratic terms behaves differently from either term alone. Let’s work it out.

   1. Write the radii for two neighbouring orbits:

   $r_n = a_0 n^2, \qquad r_{n+1} = a_0 (n+1)^2.$

   2. The gap between them is

   $\Delta r = r_{n+1} - r_n = a_0 \left[ (n+1)^2 - n^2 \right].$

   3. Expand $(n+1)^2 = n^2 + 2n + 1$. Then

   $\Delta r = a_0 \left( n^2 + 2n + 1 - n^2 \right) = a_0 (2n + 1).$

   4. For large $n$, the constant $1$ becomes negligible compared to $2n$. So

   $\Delta r \approx 2 a_0 n.$

   Thus the spacing between consecutive orbits is proportional to $n$ itself — not $n^2$, not $\sqrt{n}$, not $n^3$. …

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3. CBSE 2026Set 55/2/11 markMCQ
   Q.Two statements are given – one labelled Assertion (A) and the other Reason (R). Select the correct answer from the codes below: (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false and (R) is also false. Assertion (A) : On increasing the intensity of incident light of frequency $\nu\,(>\nu_0)$ on a photosensitive surface, the photocurrent increases. Reason (R) : The stopping potential for a photosensitive surface increases with increase of frequency $\nu\,(>\nu_0)$ of incident light.

   ›Reveal solutionSolution

   The Assertion is true because photocurrent depends on the number of photoelectrons, which increases with light intensity. The Reason is also true — stopping potential depends on frequency, not intensity. But the Reason does not explain the Assertion, since they involve different physical mechanisms. So the correct choice is (B).

   The core physics: photoelectric effect and the two separate ideas

   The photoelectric effect has a clean conceptual split that many students miss. There are two independent things happening when light hits a metal surface:

   1. How many electrons come out — this is governed by the intensity of light (number of photons per second). More photons → more electrons ejected → larger photocurrent.
   2. How much kinetic energy each electron has — this is governed by the frequency of light (energy per photon). Higher frequency → more energy per electron → you need a larger stopping potential to bring them to rest.

   These two are completely separate. Intensity does not affect the energy of individual electrons; frequency does not affect how many electrons are ejected (above threshold). This is the key insight that makes the question straightforward.

   ---

   Step-by-step reasoning

   1. Understanding the Assertion (A)

   On increasing the intensity of incident light of frequency $\nu > \nu_0$ on a photosensitive surface, the photocurrent increases.

   When frequency is above the threshold $\nu_0$, every photon that hits the surface can eject a photoelectron (assuming it's absorbed). Intensity means the number of photons per second per unit area. If you increase intensity, more photons arrive each second, so more electrons are knocked out. The photocurrent (rate of flow of charge) is directly proportional to the number of electrons emitted per second.

   Photocurrent $\propto$ Intensity (for $\nu > \nu_0$)

   So the Assertion is true.

   2. Understanding the Reason (R)

   The stopping potential for a photosensitive surface increases with increase of frequency $\nu > \nu_0$ of incident light.

   Stopping potential $V_s$ is the voltage needed to stop the most energetic photoelectrons. Einstein's photoelectric equation gives:

   $K_{\text{max}} = h\nu - \phi_0$

   where $\phi_0$ is the work function. Since $K_{\text{max}} = eV_s$, we have:

   $eV_s = h\nu - \phi_0 \quad \Rightarrow \quad V_s = \frac{h}{e}\nu - \frac{\phi_0}{e}$

   This is a straight line with slope $h/e$. As $\nu$ increases, $V_s$ increases linearly. So the Reason is also true. …

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4. CBSE 2026Set 55/3/11 markMCQ
   Q.Radiation of wavelength $200$ nm is incident on a photosensitive surface of work function $4.2$ eV. The kinetic energy of the fastest photoelectrons emitted from this surface will be close to : (A) $3.5$ eV (B) $3.0$ eV (C) $2.5$ eV (D) $2.0$ eV

   ›Reveal solutionSolution

   The fastest photoelectron’s kinetic energy is found from Einstein’s photoelectric equation: $K_{\text{max}} = h\nu - \phi$. Converting wavelength to frequency and using the given work function gives $K_{\text{max}} \approx 2.0$ eV, so the correct option is (D).

   The photoelectric effect is a clean, direct application of energy conservation: a photon gives all its energy to an electron. The electron uses some of that energy to escape the surface (the work function $\phi$), and whatever remains shows up as kinetic energy. The fastest electron is the one that loses the least energy on its way out — so its kinetic energy is simply $h\nu - \phi$.

   Let’s walk through it.

   1. Find the photon energy in eV. Wavelength $\lambda = 200$ nm $= 200 \times 10^{-9}$ m. Photon energy $E = h\nu = \frac{hc}{\lambda}$. Use the handy constant $hc = 1240$ eV·nm (this is a standard shortcut for such problems). So

   $E = \frac{1240\ \text{eV·nm}}{200\ \text{nm}} = 6.2\ \text{eV}.$

   Tip

   Memorise $hc \approx 1240$ eV·nm — it converts wavelength in nm directly to energy in eV without messing with SI units.

   2. Apply Einstein’s photoelectric equation. The maximum kinetic energy of emitted photoelectrons is

   $K_{\text{max}} = h\nu - \phi.$

   Here $\phi = 4.2$ eV. So

   $K_{\text{max}} = 6.2\ \text{eV} - 4.2\ \text{eV} = 2.0\ \text{eV}.$ …

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5. CBSE 2026Set 55/3/11 markMCQ
   Q.Assertion (A) : Photoelectric current depends upon the intensity of the incident radiation. Reason (R) : Stopping potential is independent of the intensity of the incident radiation. (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.

   ›Reveal solutionSolution

   The photoelectric current depends on intensity (more photons → more electrons), while stopping potential depends only on frequency (photon energy minus work function). Both statements are true, but the reason does not explain the assertion — they are independent facts about the photoelectric effect. The correct option is (B).

   The photoelectric effect is one of those rare topics where a single experiment cleanly separates two ideas that students often mix up: how many electrons get ejected versus how energetic each electron is. The assertion and reason here test exactly that distinction.

   The core physics: When light hits a metal surface, each photon can knock out one electron — but only if the photon’s energy ($hf$) exceeds the metal’s work function ($\phi$). The number of electrons ejected (and hence the current) depends on how many photons arrive per second — that’s intensity. But the maximum kinetic energy of each ejected electron depends only on the photon’s frequency, not on how many photons there are.

   Let’s walk through each statement carefully.

   1. Assertion (A): Photoelectric current depends on the intensity of incident radiation.

      Current is charge per second. Each ejected electron carries charge $e$. So if more electrons leave the metal per second, the current increases. What controls the number of electrons per second? The number of photons striking the metal per second — that is, the intensity (for a fixed frequency).

      Tip

      Think of it like rain: intensity is how hard it’s raining (number of raindrops per second). More raindrops → more splashes (electrons). But each splash’s height depends on the size of the raindrop (frequency), not on how many are falling.

      So (A) is true.

   2. Reason (R): Stopping potential is independent of the intensity of incident radiation.

      Stopping potential $V_0$ is the voltage needed to stop the most energetic photoelectrons. Einstein’s photoelectric equation gives:

   $eV_0 = hf - \phi$

   Here $f$ is the frequency of light, $\phi$ is the work function. Notice: intensity does not appear anywhere in this equation. Whether you shine a dim light or a bright light of the same colour, the stopping potential stays the same. …

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6. CBSE 2026Set DS1 mark
   Q.The work function of a metal is 3.3 eV. Calculate the minimum frequency of photon that will emit the photoelectron.

   ›Reveal solutionSolution

   The minimum (threshold) frequency is $\nu_0 = W/h \approx 8.0\times10^{14}$ Hz.

   Concept. Photoelectrons are emitted only when the incident photon's energy at least equals the work function $W$ of the metal. The minimum (threshold) frequency corresponds to a photon whose whole energy is used up just to free the electron, with no kinetic energy left:

   $h u_0 = W \;\Rightarrow\; u_0 = \frac{W}{h}.$

   …

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7. CBSE 2026Set A1 markMCQ
   Q.The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2

   ›Reveal solutionSolution

   Bohr radii scale as n², so r₂ = 4 r₀.

   In Bohr's model of hydrogen the radius of the $n$-th orbit is:

   $r_n=n^2 r_0$

   …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.When light falls on a metal surface, the maximum kinetic energy of the emitted electrons depends upon

   (a) the time for which light falls on the metal

   (b) the frequency of the incident light

   (c) the intensity of the incident light

   (d) the velocity of the incident light

   ›Reveal solutionSolution

   Maximum kinetic energy of photoelectrons depends only on the frequency of incident light (and the metal's work function), never on intensity or time.

   By Einstein's photoelectric equation,

   $KE_{max} = h\nu - \phi_0$

   where $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function of the metal (a fixed property of the metal surface). Increasing the intensity of light (at fixed frequency) increases the number of photoelectrons emitted per second (photocurre …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is

   (a) 9R

   (b) R/3

   (c) 3R

   (d) R/9

   ›Reveal solutionSolution

   Bohr radius of the nth orbit scales as $n^2$, so the third orbit's radius is $9R$.

   In the Bohr model, the radius of the $n$th orbit of the hydrogen atom is

   $r_n = n^2 r_1$

   …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The work function of four metals P, Q, R and S are 2.3 eV, 3.2 eV, 4.25 eV and 5.15 eV respectively. Among these the metal having lowest threshold frequency will be

    (a) R

    (b) S

    (c) Q

    (d) P

    ›Reveal solutionSolution

    Threshold frequency is directly proportional to work function (nu0 = W/h), so the metal with the smallest work function has the lowest threshold frequency.

    Einstein's photoelectric equation gives the threshold frequency as nu0 = W/h, where W is the work function and h is Planck's constant (a fixed constant). Since nu0 scales directly with W, the metal with the smallest work function has the smallest ( …

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11. CBSE 2026Set ANNUAL1 mark
    Q.If the workfunction of a metal is 4.5 eV and the kinetic energy of the photoelectron emitted from its surface is 1.5 eV, then calculate the energy of a photon of the incident light.

    ›Reveal solutionSolution

    Einstein's photoelectric equation says the photon's energy is used partly to free the electron (work function) and partly becomes its kinetic energy; adding the two gives the photon energy.

    …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.Photoelectric effect supports

    (a) wave nature of light

    (b) particle nature of light

    (c) dual nature of light

    (d) polarization nature of light

    ›Reveal solutionSolution

    Classical wave theory cannot explain the instantaneous emission or the frequency threshold seen in the photoelectric effect; only Einstein's photon (particle) picture of light does.

    In the photoelectric effect, light falling on a metal surface ejects electrons. Key observations - that emission starts INSTANTLY (no time lag) even at very low intensity, that there is a definite THRESHOLD FREQUENCY below which no electrons are emitted no matter how intense the light, and that the maximum kinetic energy of emitted electrons depends on FREQUENCY (not intensity) - cannot be explained by the classical wave theory of light. Einstein explained these features by treating light as made of discrete quanta (photons) of …

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