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) $3.978\times10^{-24}$

The momentum (in kg m/s) of a photon of frequency 6.0×10¹⁴ Hz is: (A) 6.63×10⁻²⁵ (B) 1.326×10⁻²⁷ (C) 2.652×10⁻

Question

6.0\times10^{14}

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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.

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
Apply the momentum formula

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

Substitute the values

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

Multiply the numerator first

6.63 \times 6.0 = 39.78

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

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

Divide by $3 \times 10^8$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

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 …

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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

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

…

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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 use …

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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 $E_2 = 2.5\ \text{eV}$ :

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

…

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CBSE 2025Set 55/6/11 markMCQ
$6.0\times10^{14}$
›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.

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

Apply the momentum formula

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

Substitute the values

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

Multiply the numerator first

$6.63 \times 6.0 = 39.78$ $10^{-34} \times 10^{14} = 10^{-20}$
So numerator = $39.78 \times 10^{-20}$

Divide by $3 \times 10^8$ …
Start your 14-day free trial to see the full solution →
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

…

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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 em …

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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 partic …

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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 $m_0 = 0$ . …

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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 m …

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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.

…

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

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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) $3.978\times10^{-24}$

Concept understanding — Photon Energy and the Particle Nature of Radiation

Q.The momentum (in kg m/s) of a photon of frequency 6.0×10146.0\times10^{14}6.0×1014 Hz is: (A) 6.63×10−256.63\times10^{-25}6.63×10−25 (B) 1.326×10−271.326\times10^{-27}1.326×10−27 (C) 2.652×10−262.652\times10^{-26}2.652×10−26 (D) 3.978×10−243.978\times10^{-24}3.978×10−24

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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) $3.978\times10^{-24}$

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