Why is Bremsstrahlung produced during electron-nucleus encounters?

[wumple]

It is my understanding that if an electron decelerates during an encounter with a 'heavy target nucleus', the energy lost by the electron is converted to an x-ray.

The book I'm using says that 'the target nucleus is so massive that the energy it acquires during the collision can safely be neglected'. Why is this? Why doesn't the nucleus take the energy that the electron lost?

 

[granpa]

consider shooting a bullet into a watermelon.
where does the energy go?

 

[marcusl]

A proton or neutron is 1836 times more massive than an electron, and a nucleus has many (often hundreds) of nucleons. Look up the mass energy of a proton, and compare it to an incoming electron with kinetic energy of, for instance, some 10's of keV (typical Bremsstrahlung experimental energy). You'll see why the absorbed energy can be neglected.

 

[granpa]

consider shooting a ping pong ball onto a watermelon.
where does the energy go?

 

[wumple]

A proton or neutron is 1836 times more massive than an electron, and a nucleus has many (often hundreds) of nucleons. Look up the mass energy of a proton, and compare it to an incoming electron with kinetic energy of, for instance, some 10's of keV (typical Bremsstrahlung experimental energy). You'll see why the absorbed energy can be neglected.

I'm still confused - I understand that the nucleus's energy is negligibly affected by the electron, but if the nucleus absorbs the electron's energy (even though it is negligible), then where does the energy for the ejected photon come from?

 

[wumple]

consider shooting a bullet into a watermelon.
where does the energy go?

Into the flying watermelon pieces and the now slower bullet?

 

[marcusl]

The electron's kinetic energy is transferred to radiation as it is accelerated (decelerated?) by the nucleus.