What is the de Broglie wavelength of a falling neutron in a vacuum?

AI Thread Summary
The discussion centers on calculating the de Broglie wavelength of a neutron that has fallen 0.60 meters in a vacuum. Participants highlight the importance of determining the neutron's velocity, which can be derived from its change in kinetic energy due to free fall. The correct formula for de Broglie wavelength is λ=h/mv, where h is Planck's constant and m is the mass of the neutron. After calculating the velocity as 3.43 m/s, the wavelength was initially miscalculated due to an incorrect value of Planck's constant. The correct calculations yield a de Broglie wavelength of approximately 5.87 x 10^-8 meters.
DRC12
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Homework Statement


What is the de Broglie wavelength of a neutron that has fallen 0.60 in a vacuum chamber, starting from rest?

Homework Equations


Not entirely sure
λ=h/mv

The Attempt at a Solution


I can't figure out what the velocity is, I think it would have to do with the change in kinetic energy but I can't figure out what that would be
 
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DRC12 said:

Homework Statement


What is the de Broglie wavelength of a neutron that has fallen 0.60 in a vacuum chamber, starting from rest?

Homework Equations


Not entirely sure
λ=h/mv

The Attempt at a Solution


I can't figure out what the velocity is, I think it would have to do with the change in kinetic energy but I can't figure out what that would be
Is the distance 0.6 m, cm, mm? Units, as well as magnitude, are rather important to a physics or engineering, or other quantitative problem.

If the neutron is in freefall - what is the acceleration? Knowing the acceleration and distance, one can determine the velocity at that distance.
 
The other way to look at it, instead of calculating accelerations and integrating back to get velocity, is to look at the Energy. The change in energy is just mg*(change in height) and this goes into the kinetic energy i.e. 1/2 mv^2.
 
Astronuc said:
Is the distance 0.6 m, cm, mm?
meters

I tried finding the velocity using v2=2ax
and found v to be 3.43m/s
then I plugged that into λ=h/mv
where m=1.67*10-27kg and h=6.67*10-34
and found λ=5.87*10-8
 
wait for some reason I was using h=3.36E-34 instead of 6.67E-34 I don't even know where that number came from
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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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