Wavelength and linear momentum

[mss90]

Homework Statement

I had to find wavelenght and linear momenta of fotons with energies of 3eV, 50 KeV and 1.0 MeV

Are these correct?

Homework Equations

E=hc/λóλ=hc/E and p= h/ λ

The Attempt at a Solution

a. 3eV Hence λ=(6.63E-34*3E8)/3=6.63E-26m
p = 6.63E-34/6.63E-26 = 1E-8
b. 50 KeV = 50000 eVλ=(6.63E-34*3E8)/ 50000 =3.978E-30m
p = 6.63E-34/3.978E-30 = 1.66E-4
c. 1.0 MeV = 1 000000 eVλ=(6.63E-34*3E8)/ 1 000000 = 1.989E-31m
p = 6.63E-34/1.989E-31 = 0.0033
[/B]

 

[collinsmark]

Hello mss90,

Welcome to PF! :)

Don't forget to first convert the energy (given in units of eV, keV, and MeV) to units of Joules first, before plugging in the numbers.

 

[mss90]

Alright, can you confirm that this is correct:

3eV * 1.60E-19 = 4.8E-19 J Hence λ=(6.63E-34*3E8)/4.8E-19=4.17E-7m = 0.417µm
p = 6.63E-34/4.17E-7= 1.59E-27

 

[collinsmark]

Alright, can you confirm that this is correct:

3eV * 1.60E-19 = 4.8E-19 J Hence λ=(6.63E-34*3E8)/4.8E-19=4.17E-7m = 0.417µm
p = 6.63E-34/4.17E-7= 1.59E-27

That looks about right, although there might be some rounding errors going on somewhere.

By the way, when calculating the photons' momentum magnitude, you can simply use the $p = \frac{E}{c}$ formula (after converting the energy into units of Joules, simply divide that by the speed of light, 3 × 10⁸ m/s, and you have the magnitude of the photon's momentum. That way you don't need to depend on the λ intermediate step as part of the answer).

 

[HalfLostAndSearching]

Your calculations for the wavelengths and linear momenta seem to be correct based on the given equations. However, it would be helpful to include the units for the final answers (i.e. meters for wavelength and kg*m/s for linear momentum). Additionally, it would be good to explain the significance of these values and how they relate to the concept of wavelength and linear momentum in the context of photons. Overall, your response is well-organized and shows a good understanding of the topic. Keep up the good work!