Q.What will be the energy of each photon in monochromatic light of frequency $5\times10^{14}$ Hz?

Showing the 12 most recent of 22 on this concept.

1. CBSE 2026Set V11 mark
   Q.In interaction with matter, light behaves as if it is made up of packets of energy called __________. Fill in the blank choosing the appropriate answer from the bracket: (photons, diffraction, polarity, monopoles, greater than unity, less than unity)

   ›Reveal solutionSolution

   photons

   ✓Final answer

   photons

   When light interacts with matter (as in the photoelectric effect or Compton effect), it behaves as if its energy is concentrated in discrete packets called photons, each carrying energy $E = h\nu$ and momentum $p = h/\lambda$.

2. CBSE 2026Set A1 markMCQ
   Q.The momentum (p) of photon is (A) λ/h (B) h/λ (C) hc/λ (D) hλ

   ›Reveal solutionSolution

   A photon's momentum is p = h/λ = E/c.

   A photon of wavelength $\lambda$ has energy $E=h\nu=\dfrac{hc}{\lambda}$. Since a photon travels at speed $c$, its momentum is

   $p=\frac{E}{c}=\frac{hc/\lambda}{c}=\frac{h}{\lambda}.$

   This is the same de Broglie relation $\lambda=h/p$ rearranged. So the photon momentum is $p=h/\lambda$.

   ✓Final answer

   (B) h/λ.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): The momentum of photon is p = h/λ. Reason (R): A photon act as a massless particle.

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is true, but Reason (R) is false.

   (d) Assertion (A) and Reason (R) both are false.

   ›Reveal solutionSolution

   A photon's momentum $p=h/\lambda$ follows directly from treating it as a massless particle with $E=pc$.

   For a massless particle, relativistic energy is $E = pc$ (since $E^2 = p^2c^2 + m^2c^4$ and $m=0$). A photon's energy is also $E = h\nu = \dfrac{hc}{\lambda}$. Equating: $pc = \dfrac{hc}{\lambda} \Rightarrow p = \dfrac{h}{\lambda}$. This derivation explicitly uses the photon's masslessness, so Reason (R) correctly explains Assertion (A).

   ✓Final answer

   Both A and R are true and R is the correct explanation of A (option a).

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.A metal surface is illuminated by photons of energy 5 eV and 2.5 eV respectively. The ratio of their wavelengths of emitted radiation is ______.

   (a) 1 : 4

   (b) 1 : 2

   (c) 2 : 1

   (d) 4 : 1

   ›Reveal solutionSolution

   Photon energy and wavelength are inversely related ($E = hc/\lambda$), so the ratio of wavelengths is the inverse of the ratio of energies.

   For a photon, $E = \dfrac{hc}{\lambda} \implies \lambda = \dfrac{hc}{E}$.

   For the two photons of energy $E_1 = 5\ \text{eV}$ and $E_2 = 2.5\ \text{eV}$:

   $\frac{\lambda_1}{\lambda_2} = \frac{hc/E_1}{hc/E_2} = \frac{E_2}{E_1} = \frac{2.5}{5} = \frac{1}{2}$

   So $\lambda_1 : \lambda_2 = 1 : 2$ — the higher-energy (5 eV) photon has the shorter wavelength, exactly half that of the 2.5 eV photon's wavelength.

   ✓Final answer

   Option (b) 1 : 2.

5. CBSE 2025Set 55/6/11 markMCQ
   Q.The momentum (in kg m/s) of a photon of frequency $6.0\times10^{14}$ Hz is: (A) $6.63\times10^{-25}$ (B) $1.326\times10^{-27}$ (C) $2.652\times10^{-26}$ (D) $3.978\times10^{-24}$

   ›Reveal solutionSolution

   The momentum of a photon is given by $p = \frac{h}{\lambda} = \frac{h\nu}{c}$. For $\nu = 6.0 \times 10^{14}$ Hz, using $h = 6.63 \times 10^{-34}$ J·s and $c = 3 \times 10^8$ m/s, the momentum is $1.326 \times 10^{-27}$ kg m/s, which matches option (B).

   The key idea here is that a photon, though massless, carries momentum. This is a purely quantum concept — you can't derive it from classical physics. The momentum of a photon is directly tied to its wave properties: the shorter the wavelength (or higher the frequency), the greater the momentum.

   The formula you need is:

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

   where $h$ is Planck's constant ($6.63 \times 10^{-34}$ J·s), $\nu$ is the frequency, and $c$ is the speed of light ($3 \times 10^8$ m/s). This relation comes from combining $E = h\nu$ (photon energy) with $E = pc$ (energy-momentum relation for massless particles).

   Now let's work through the calculation step by step.

   1. Write down the given data

      Frequency, $\nu = 6.0 \times 10^{14}$ Hz

      Planck's constant, $h = 6.63 \times 10^{-34}$ J·s

      Speed of light, $c = 3 \times 10^8$ m/s

   2. Apply the momentum formula

   $p = \frac{h\nu}{c}$

   3. Substitute the values

   $p = \frac{(6.63 \times 10^{-34}) \times (6.0 \times 10^{14})}{3 \times 10^8}$

   4. Multiply the numerator first

   $6.63 \times 6.0 = 39.78$

   $10^{-34} \times 10^{14} = 10^{-20}$

   So numerator = $39.78 \times 10^{-20}$

   5. Divide by $3 \times 10^8$

   $\frac{39.78 \times 10^{-20}}{3 \times 10^8} = \frac{39.78}{3} \times 10^{-20-8} = 13.26 \times 10^{-28}$

   6. Write in proper scientific notation

   $13.26 \times 10^{-28} = 1.326 \times 10^{-27}$

   Watch out

   A common mistake is to forget that $10^{-20}$ divided by $10^8$ gives $10^{-28}$, not $10^{-12}$. Always subtract exponents carefully when dividing powers of ten.

   Tip

   You can also think of this as: first find the wavelength $\lambda = c/\nu = 5 \times 10^{-7}$ m (500 nm, which is green light), then use $p = h/\lambda$. Either path gives the same result.

   ✓Final answer

   The momentum of the photon is $1.326 \times 10^{-27}$ kg m/s, which corresponds to option (B).

6. CBSE 2025Set D1 markMCQ
   Q.The formula of kinetic mass of photon is (A) hν/c (B) hν/c^2 (C) hc/ν (D) c^2/hν

   ›Reveal solutionSolution

   Equate the photon energy hν with mc²; solving gives m = hν/c².

   A photon has energy E = hν. By mass–energy equivalence, an energy E corresponds to an effective mass m through E = mc².

   Setting the two equal:

   hν = mc² ⟹ m = hν/c²

   This is the photon's kinetic (relativistic) mass; the photon has zero rest mass. Equivalently, using ν = c/λ, m = h/(λc).

   ✓Final answer

   (B) hν/c².

7. CBSE 2025Set A1 mark
   Q.Match Column 'A' item 'Intensity 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

   Intensity of light corresponds to option (iv): the number of photons.

   In Einstein's photon picture, the intensity of a beam of monochromatic light is determined by how many photons strike a given area per unit time (not by the energy of each individual photon, which instead depends only on frequency). Doubling the intensity of light of a fixed frequency doubles the number of photons per second, which in the photoelectric effect increases the photoelectric current (rate of electron emission) without changing the maximum kinetic energy of each emitted electron. Hence 'Intensity of light' is correctly paired with '(iv) Number of photons'.

   ✓Final answer

   Intensity of light → (iv) Number of photons.

8. CBSE 2025Set A1 mark
   Q.Match Column 'A' item 'Particle nature 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

   The particle nature of light corresponds to option (vi): the photon.

   While phenomena like interference, diffraction and polarisation reveal light's wave nature, the photoelectric effect (and Compton effect) reveal that light also behaves as a stream of discrete, localised energy packets called photons, each carrying energy $E = h\nu$ and momentum $p = h/\lambda$, and interacting with matter (like an electron) as a single indivisible unit — just like a particle. This particle-like behaviour of light is captured entirely by the concept of the photon. Hence 'Particle nature of light' matches '(vi) Photon'.

   ✓Final answer

   Particle nature of light → (vi) Photon.

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The rest mass of photon is :

   (a) 1 kg

   (b) Infinite

   (c) 1 g

   (d) zero

   ›Reveal solutionSolution

   A photon always travels at speed c and carries energy/momentum without ever being at rest, so its rest mass must be zero.

   A photon is a quantum of electromagnetic energy. Its energy and momentum are related by $E = pc$ (from relativity, valid for a massless particle), and separately $E = h\nu$. Applying the relativistic energy–momentum relation

   $E^2 = p^2c^2 + m_0^2c^4$

   for a photon with $E = pc$, this forces $m_0^2c^4 = 0$, i.e. the rest mass $m_0 = 0$.

   Physically, a photon can never be brought to rest — it always moves at speed $c$ in vacuum — so the concept of a nonzero "rest mass" does not apply; it is exactly zero.

   ✓Final answer

   Zero — option (d).

10. CBSE 2022Set ANNUAL1 markMCQ
    Q.Photons are electrically ____.

    (a) neutral

    (b) positive

    (c) negative

    (d) unpredictable

    ›Reveal solutionSolution

    A photon carries energy and momentum but no electric charge.

    A photon is the quantum (packet) of electromagnetic radiation. It carries energy $E = h\nu$ and momentum $p = h\nu/c$, but it has zero rest mass and zero electric charge. Because photons are uncharged, they are not deflected by electric or magnetic fields, unlike charged particles such as electrons or protons.

    ✓Final answer

    Photons are electrically neutral — option (a).

11. CBSE 2022Set TERM21 markMCQ
    Q.Momentum of Photon of frequency 'v' is:

    (a) zero

    (b) hv/c

    (c) hc/v

    (d) 2hc/v

    ›Reveal solutionSolution

    A photon's momentum follows from $E=h\nu$ and the relativistic relation $E=pc$ for a massless particle.

    A photon has energy $E = h\nu$. Since a photon is massless and travels at speed $c$, its energy and momentum are related by $E = pc$. Therefore:

    $p = \frac{E}{c} = \frac{h\nu}{c}$

    ✓Final answer

    $p = h\nu/c$. Correct option: (b).

12. CBSE 2021Set A1 markMCQ
    Q.The rest mass of photon is (A) zero (B) infinite (C) 9.1 × 10⁻³¹ kg (D) 1.6 × 10⁻²⁷ kg

    ›Reveal solutionSolution

    The rest mass of a photon is zero.

    A photon always moves with the speed of light c. From relativity, any particle with non-zero rest mass would require infinite energy to reach the speed c. Since a photon actually moves at c and carries finite energy E = hν and momentum p = E/c = h/λ, its rest mass must be exactly zero. (Its relativistic/effective mass hν/c² is non-zero, but its rest mass is zero.)

    ✓Final answer

    (A) zero.