Momentum of a proton in de Broglie wavelength

AI Thread Summary
The de Broglie wavelength of a proton is derived from its momentum, expressed as λ=h/p. When protons are accelerated from rest through a potential V, their momentum can be related to the energy gained, leading to the equation mv²/2 = eV. By substituting this expression for velocity into the de Broglie wavelength formula, the correct answer is determined to be B) h/√(2meV). The discussion emphasizes the relationship between energy, momentum, and wavelength in quantum mechanics. Understanding these concepts is crucial for solving problems involving particle behavior at the quantum level.
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Homework Statement



In quantum mechanics the de Broglie wavelength of an object depends
on its momentum according to λ=h/p where h is Planck's constant.
Protons of charge e and mass m are accelerated from rest through a
potential V. What is their de Broglie wavelength?

A) 2h/\sqrt{}meV
B) h/\sqrt{}2meV
C) h\sqrt{}meV
D) h/eV

Homework Equations



I don't know.

The Attempt at a Solution



I just don't know what I have to search for to solve it.
How can I express p with e and V?
 
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Hello there.
As you probably know p=mv. From conservation of energy u have mv^2/2 = eU, here you can solve for the velocity v. After that you can use it to end up with answer B)
 
Thanks.
 
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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