Buckyball Diffraction Thru a Grating.

[TMO]

Homework Statement

Buckyballs are soccerball-shaped molecules consisting of 60 carbon atoms (Mass of C₆₀ = 1.2024×10^-24 kg) with an approximate diameter of 1 nm. A beam of buckyballs with each molecule carrying a kinetic energy of 0.60 eV is normally incident on a grating with a slit with of 10 nm and 10⁵ lines per centimeter. We detect these molecules with suitable equipment placed at a distance of 1.5 m behind the grating. Let the wavelength of these buckyballs be λ.

(1) How far do we have to move the detector from the path of incidence to find the first maximum of intensity?

(2) If a photon had the same energy E as a buckyball of momentum p, what would be its wavelength λ_φ? Give an algebraic answer in terms of m_C60, h, and c only.

Homework Equations

(I) d sin(θ) = ±mλ, where m is an integer.
(II) p²/(2m) = E
(III) p_φ = E/c
(IV) E = hc/λ_φ

The Attempt at a Solution

For (1), the maximae occur at integral multiples of the wavelength, so the detector should be moved λ wavelengths away (or so I think).

For (2), cross multiply equation (III) to get E = pc. Set this equal to (I) yielding p = 2m_C60c. Plug equation (IV) into (III) yielding p = h/λ_φ. Therefore 2m_C60c = h/λ_φ. Trivial algebra gives the desired solution λ_φ = h/(2m_C60c)

My solution to problem (1) feels too simple to be right, and my solution to problem (2) assumes that p = p_φ, and I'm unsure as to whether such an equivalence is correct. Any help?

 

[Simon Bridge]

For (1), the maximae occur at integral multiples of the wavelength, so the detector should be moved λ wavelengths away (or so I think).

Nope - the path difference has to be an integer number of wavelengths. That's why ##d\sin\theta=m\lambda##

 

[TMO]

Okay then. Let ρ be the line density per meter. Then the distance between each line should be given by ρ^-1. Thus ρ^-1 sin(θ) ≈ ρ^-1ΔyR^-1 = λ. ∴ Δy = λρR.

 

[Simon Bridge]

Why not just use the deBroglie wavelength of the buckyball and proceed as for optics?

 

[TMO]

Why not just use the deBroglie wavelength of the buckyball and proceed as for optics?

When I said in my original post, "Let the wavelength of these buckyballs be λ," I was referring to the deBroglie wavelength of the buckyball. Furthermore, I am referring to the same lambda in the following quoted post as in the original post:

Okay then. Let ρ be the line density per meter. Then the distance between each line should be given by ρ^-1. Thus ρ^-1 sin(θ) ≈ ρ^-1ΔyR^-1 = λ. ∴ Δy = λρR.

 

[Simon Bridge]

Oh excuse me - I misread the relations.
Note: people marking exams can do that too - which is why you annotate your working.

Then what's the problem?