Answer: Understanding Electron Volt Energy for X-Ray Machines

AI Thread Summary
X-ray energies are measured in electron volts (eV), where 1 eV equals 1.601 x 10^-19 joules. In an x-ray machine, an electron with a kinetic energy of 10 keV can produce x-ray photons with a maximum energy equal to that kinetic energy. The concept of energy conservation is crucial in understanding this interaction. The discussion highlights a realization about the relationship between kinetic energy and photon energy. Understanding these principles is essential for solving related problems in x-ray physics.
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Homework Statement



Hi guys.

X-ray energies are commonly given i nunits of "electron volts". 1 eV is the energy that an electron would have after being accelerated across a potential different of 1V and is equivalent to 1.601 *10^-19
In an x-ray machine an electron that is about to hit an electrode has a kinetic energy of 10keV. What is the maximum energy of the x-ray photon that this interaction could produce?

Homework Equations





The Attempt at a Solution



I don't even know where to start.
 
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Just consider energy conservation.

If I give you 10 apples (and you have no apples to begin with), what is the maximal number of apples you can give me back?
 
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damn.Hahhaha thanks man. I can't believe i never realized that.
 
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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