Question

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

Answer 1

To help you I need to assume a wavelength and then calculate the momentum.

The momentum equation for photons is:

p = h / λ , this is the division of Plank's constant by the wavelength.

Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:

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

p = 1.01067 * 10^ - 27 kg*m/s which  must be rounded to three significant figures.

With that, p = 1.01 * 10 ^ -27 kg*m/s

The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s

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

Answer 2

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

Further Explanation:

According to the hypothesis given by the scientist De-Broglie, every material particle is associated with a wave called as matter wave and the wavelength of the particle is related to the particle’s momentum as follows:

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

Here, $h$ is the Planck’s constant, $p$ is the momentum of the particle, $\lambda$ is the wavelength of the matter wave associated with the moving particle.

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

Convert the wavelength of the photon associated with the wave into meter.

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

Rearrange the above equation for the momentum of the photon.

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

Now, substituting the values of Planck’s constant and the wavelength of the photon.

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

 

Hence, momentum of the matter wave associated with the photon is  $\boxed{1.0 \times {10^{ - 27}}{{{\text{ kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ by considering the wavelength to be $650\text{ nm}$.

Learn More:

1. Suppose a 60kg gymnast climbs a rope brainly.com/question/4124873

2. Which of the following about electromagnetic radiations is true brainly.com/question/1619496

3. The results of Rutherford's gold foil experiment brainly.com/question/1542931

Answer Details:

Grade: College

Subject: Physics

Chapter: Modern Physics

Keywords:

Momentum, photon, week's note, wavelength, plank's constant, 650nm, de-broglie, light, visible, p=h/lambda, 6.646x10^-34, hypothesis, moving particle.