Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Plank’s constant (6.63E-34

Question

kondaur

19 hours ago

14

Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Plank’s constant (6.63E-34

) to calculate the momentum of this photon:
1.0E-27 kgm/s

1.8E-27 kgm/s

2.0E-27 kgm/s

3.0E-27 kgm/s

Physics

2 answers:

Yuliya22 [1.1K] · Answer 1 · 2026-03-05T01:20:34+03:00

Option 1 is correct. The momentum associated with the photon of the wave is $\boxed{1.0 \times {10^{ - 27}}{{{\text{ kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ , considering the wavelength as $650\text{ nm}$ .

Additional Explanation:

According to De-Broglie's hypothesis, every matter particle is linked to a matter wave, with the wavelength of that particle directly linked to its momentum as follows:

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

Where $h$ represents Planck’s constant, $p$ indicates the particle's momentum, and $\lambda$ denotes the wavelength of the associated matter wave for the moving particle.

The value for Planck's Constant is $h = 6.646 \times {10^{ - 34}}{\text{ J}\cdot{s}}$ .

Convert the photon's wavelength associated with this wave into meters.

$\begin{aligned}\lambda&=650\text{ nm}\\&=650\times10^{-9}\text{ m}\\&=6.50\times10^{-7}\text{ m}\end{aligned}$

Rearranging the above equation allows us to solve for the photon's momentum.

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

Next, substitute in the values for Planck’s constant and the wavelength of the photon.

$\begin{aligned}p&=\dfrac{{6.646\times{{10}^{-34}}{\text{ J.s}}}}{{{\text{650}} \times {\text{1}}{{\text{0}}^{-9}}{\text{ m}}}}\\&=1.022\times{10^{-27}}{\text{ kg.m/s}}\\&=1.0\times {10^{-27}}{\text{kg.m/s}}\\\end{aligned}$

Thus, the momentum corresponding with the matter wave of the photon becomes $\boxed{1.0 \times {10^{ - 27}}{{{\text{ kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ , while considering the wavelength as $650\text{ nm}$ .

Further Learning:

1. Suppose a 60kg gymnast climbs a rope

2. What are the true statements concerning electromagnetic radiation?

3. Results of Rutherford's gold foil experiment

Response Information:

Grade Level: College

Subject: Physics

Chapter: Modern Physics

Keywords:

Momentum, photon, weekly notes, wavelength, Planck's constant, 650nm, de-Broglie, light, visible, p=h/lambda, 6.646x10^-34, hypothesis, moving particle.

ValentinkaMS [1.1K] · Answer 2 · 2026-03-05T01:21:47+03:00

In order to assist you, I’ll need to take a wavelength as a given value and subsequently compute the momentum.

The photon momentum can be calculated using the equation:

p = h / λ, which means Plank's constant divided by the wavelength.

Let’s assume λ = 656 nm = 656 * 10 ^ - 9 m, a value derived from a previous question, leading to:

p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s

This simplifies to p = 1.01067 * 10^ - 27 kg*m/s, which we need to express rounded to three significant figures.

So we have p = 1.01 * 10 ^ -27 kg*m/s

Rounding to only two significant figures, we obtain 1.0 * 10 ^ - 27 kg*m/s

Thus, assuming the wavelength λ is 656 nm, the first option correctly estimates the answer: 1.0*10^-27 kg*m/s.