De Broglie wavelength calculations

AI Thread Summary
The calculations for the de Broglie wavelength of an alpha particle traveling at 2x10^6 m/s appear correct, yielding a wavelength of approximately 4.99x10^-14 m. The mass of the alpha particle is consistently 6.645x10^-27 kg unless stated otherwise. The relativistic effects on mass become negligible at low speeds, which is why some guides omit the square root term in calculations. Larger bodies exhibit negligible wave-like properties due to their significantly smaller de Broglie wavelengths compared to their size, making detection impractical. The discussion highlights the contrast between atomic and macroscopic scales in observing wave-particle duality.
rugapark
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I had a go at 2 Q's and wanted to make sure I'm doing this right.
so here's the first one, and maybe if i went wrong with it I was going to redo the 2nd Q on my own.

\lambda = h/p = h/mv (\sqrt{1-((v^2)/(c^2))})

so, an Alpha Particle traveling @ 2x106m/s (mass = 6.645x10-27 kg)

\lambda = [(6.626*10^-^3^4)/((6.645*10^-^2^7)*(2*10^6))] x {\sqrt{1-[(2*10^6)^2/(3*10^8)^2]}}

= (4.986x10-14) x (99.998x10-2)

= 4.99x10-14m

how does this look?
and also, am i right in assuming if the question does not state otherwise, that the mass of an alpha particle is always 6.645x10-27kg?

Cheers guys
Ruga
 
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That looks right. A particle's mass would increase as the particle's speed approaches the speed of light, so the wavelength decreases.

The rest mass of the alpha particle is always in its inertial frame 3727 MeV or 6.645 x 10-27 kg. In classical mechanics, particle mass usually refers to rest mass.
 
brilliant - just another quicky, some of the answer guides ignored the whole square root part of the equation.. is that because that part of the equation is always roughly equal to 1?
 
At low speeds, e.g. v = 0.01 c, then (v/c)2 = 0.0001, and the square root of 1-(0.01)2 = 0.99995, so the relativistic effect is very small.

Alpha particles coming from alpha decay or in fusion reactions have kinetic energies on the order of several MeV, so there speeds are not relativistic.
 
I ended up discussing something with friends when we were going through this question - why is it that we don't see wave like properties in larger bodies i.e. in macroscopic levels? is it because the larger the mass, the smaller the de Broglie wavelength, and so the wave like properties are just too small to be detected?
 
rugapark said:
I ended up discussing something with friends when we were going through this question - why is it that we don't see wave like properties in larger bodies i.e. in macroscopic levels? is it because the larger the mass, the smaller the de Broglie wavelength, and so the wave like properties are just too small to be detected?
Compare the 'size' of an alpha particle (or atomic nucleus) with the deBroglie wavelength (in the OP), then compare the wavelength of a 1 kg metal sphere (density = 8 g/cm3) with the deBroglie wavelength for different speeds, e.g. 10 m/s and 1000 m/s.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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