Q.Match Column 'A' item 'Frequency of light' with the correct option from Column 'B' and write the correct pair. Column 'B' options:

Showing the 12 most recent of 30 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.If the wavelength of a photon is halved, then its frequency will become ______.

   ›Reveal solutionSolution

   Since nu = c/lambda for a photon, halving lambda directly doubles nu (c is a universal constant).

   A photon's frequency and wavelength are related by nu = c/lambda, where c (speed of light) is fixed. If lambda is halved (lambda -> lambda/2), then nu = c/(lambda/2) …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The mass of a photon is:

   (a) h/v

   (b) hc/λ

   (c) h/λ

   (d) hν/c²

   ›Reveal solutionSolution

   A photon's energy is $E=h\nu$; equating this to $E=mc^2$ gives its (relativistic/effective) mass $m=h\nu/c^2$.

   A photon has zero rest mass but carries energy $E=h\nu$ (Planck's relation) and momentum $p=h/\lambda=E/c$. Using Einstein's mass-energy equivalence $E=mc^2$ for the energy it carries while in motion, its effective mass is $m=\dfrac{E}{c^2}=\dfrac{h\nu}{c^2}$. The other options are q …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: The rest mass of photon is ______.

   ›Reveal solutionSolution

   A photon's rest mass is zero.

   A photon is a quantum of electromagnetic radiation that always travels at the speed of light c in vacuum. According to relativity, any particle moving at speed c must have zero rest mass; otherwise its energy would be infinite. A photon does …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: The value of Planck's constant is ______.

   ›Reveal solutionSolution

   Planck's constant h ≈ 6.63 × 10⁻³⁴ J·s.

   Planck's constant h relates the energy of a photon to its frequency by E = hν. Its accepted value is

   h = 6.63 × 10⁻³⁴ joule-second (J·s).

   …

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5. CBSE 2025Set 55/4/11 markMCQ
   Q.A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true? (A) The blue beam has more number of photons than the red beam. (B) The red beam has more number of photons than the blue beam. (C) Wavelength of red light is lesser than the wavelength of blue light. (D) The blue light beam has lesser energy per photon than that in the red light beam.

   ›Reveal solutionSolution

   Since blue photons carry more energy than red photons, equal-intensity beams require more red photons to match the same total power. The red beam has more photons.

   The key to this problem lies in understanding what "intensity" means and how photon energy depends on wavelength.

   Intensity measures the power (energy per unit time) delivered per unit area. When two beams have equal intensities, they carry the same total energy per second through the same cross-sectional area, regardless of color.

   Each photon carries energy $E = h\nu = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Blue light has a shorter wavelength than red light ($\lambda_{\text{blue}} < \lambda_{\text{red}}$), which means blue photons are individually more energetic than red photons.

   If the total power delivered by both beams is the same, but blue photons pack more energy each, then fewer blue photons are needed to deliver that power. Conversely, more red photons are required to compensate for their lower individual energy.

   Let me work through this quantitatively:

   1. Express intensity in terms of photon count. If $n$ photons pass through area $A$ in time $t$, the intensity is:

   $I = \frac{\text{Total energy}}{A \cdot t} = \frac{n \cdot E_{\text{photon}}}{A \cdot t} = \frac{n \cdot hc/\lambda}{A \cdot t}$

   2. Set up the equal-intensity condition. For red and blue beams with equal intensities:

   $I_{\text{red}} = I_{\text{blue}}$

   $\frac{n_{\text{red}} \cdot hc/\lambda_{\text{red}}}{A \cdot t} = \frac{n_{\text{blue}} \cdot hc/\lambda_{\text{blue}}}{A \cdot t}$

   3. Simplify to find the photon ratio:

   $n_{\text{red}} \cdot \frac{1}{\lambda_{\text{red}}} = n_{\text{blue}} \cdot \frac{1}{\lambda_{\text{blue}}}$

   $\frac{n_{\text{red}}}{n_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}}$ …

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6. CBSE 2025Set 55/5/11 markMCQ
   Q.Which of the following electromagnetic waves has photons of the largest momentum? (A) X-rays (B) AM radio waves (C) Microwaves (D) TV waves

   ›Reveal solutionSolution

   Photon momentum is $p = \frac{h}{\lambda}$, so the wave with the shortest wavelength has the largest momentum. Among the options, X-rays have the shortest wavelength, hence the largest photon momentum.

   Concept & Intuition

   The momentum of a photon is not like the momentum of a massive particle. For a photon, momentum is purely a wave property, given by the de Broglie relation:

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

   where $h$ is Planck’s constant and $\lambda$ is the wavelength. This means: shorter wavelength → larger momentum. There is no dependence on amplitude or intensity — only wavelength matters.

   So the question reduces to: which of these electromagnetic waves has the shortest wavelength? Let’s recall the electromagnetic spectrum order from longest to shortest wavelength:

   1. Radio waves (including AM and TV) — longest wavelengths (metres to kilometres)
   2. Microwaves — centimetres to millimetres
   3. Infrared — micrometres
   4. Visible light — hundreds of nanometres
   5. Ultraviolet — tens of nanometres
   6. X-rays — picometres to nanometres
   7. Gamma rays — sub-picometre

   Watch out

   A common mistake is to think that higher frequency means higher energy (true), but then incorrectly assume that momentum depends on something else like the wave’s “penetrating power” or “ionising ability”. Stick to $p = h/\lambda$ — it’s the only formula that matters here.

   Step-by-step solution

   1. Write the momentum formula

      For any photon, $p = \frac{h}{\lambda}$. Since $h$ is constant, $p \propto \frac{1}{\lambda}$.

   2. Identify the wavelengths of each option

      • AM radio waves: wavelength $\approx 100$ m to $1000$ m (longest) …

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7. CBSE 2025Set D1 markMCQ
   Q.What is the energy of a photon with a wavelength of 500 nm? ( Use c = 3 × 10^8 m/s and h = 6.626 × 10^-34 Js ) (A) 4 × 10^-19 J (B) 2.5 × 10^-19 J (C) 1.2 × 10^-18 J (D) 6.6 × 10^-19 J

   ›Reveal solutionSolution

   Photon energy E = hc/λ ≈ 4 × 10⁻¹⁹ J for λ = 500 nm.

   The energy of a photon is

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

   Substitute h = 6.626×10⁻³⁴ J·s, c = 3×10⁸ m/s, λ = 500 nm = 500×10⁻⁹ m:

   $E = \frac{(6.626\times10^{-34})(3\times10^{8})}{500\times10^{-9}}$ …

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8. CBSE 2025Set A1 mark
   Q.Match Column 'A' item 'Frequency of light' with the correct option from Column 'B' and write the correct pair. Column 'B' options:

   (i) Minimum energy to emit electrons from the surface

   (ii) Minimum frequency to emit electrons from the surface

   (iii) Frequency of photon

   (iv) Number of photons

   (v) Moving particle

   (vi) Photon

   (vii) Einstein.

   ›Reveal solutionSolution

   Frequency of light corresponds to option (iii): frequency of photon.

   In the photon (particle) picture of light proposed by Einstein, a beam of light of frequency ν is regarded as a stream of photons, each carrying energy E = hν — so the 'frequency of light' (a wave concept) and the 'frequency of the photon' (used to compute each photon's quantum of energy) are simply the same physical quantity viewed from the t …

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9. CBSE 2025Set ANNUAL1 mark
   Q.Electron volt (eV) is the unit of ................. (fill in the blank)

   ›Reveal solutionSolution

   The electron volt is a convenient small unit of energy, widely used in atomic and nuclear physics.

   One electron volt is defined as the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt:

   $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ …

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10. CBSE 2025Set ANNUAL1 mark
    Q.A blue lamp mainly emits light of wavelength $4500\overset{\circ}{A}$. The lamp is rated at 150 W and 8% of energy is emitted as visible light. How many photons are emitted by lamp per second?

    ›Reveal solutionSolution

    Visible-light power = 8% of 150 W; divide by the energy of one photon at 4500 Å.

    Power emitted as visible light $= 8\%$ of $150\,$W $= 0.08\times150 = 12\,$W.

    Energy of one photon at $\lambda=4500\,\text{Å}=4.5\times10^{-7}\,$m:

    $E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^8}{4.5\times10^{-7}} \approx 4.42\times10^{-19}\ \text{J}$

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.The momentum of a photon of energy h.nu is

    (i) h.nu

    (ii) h.nu/c

    (iii) h.nu.c

    (iv) h/nu

    ›Reveal solutionSolution

    Photon momentum p = E/c = h(nu)/c.

    …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The momentum (p) of photon is -

    (a) h/λ

    (b) λ/h

    (c) hC/λ

    (d) hλ

    ›Reveal solutionSolution

    A photon's momentum follows from combining its energy E = hc/λ with the relativistic relation E = pc for a massless particle.

    A photon of frequency ν has energy $E = h\nu = \dfrac{hc}{\lambda}$ (since $c = \nu\lambda$).

    …

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