Question

What is the speed of an electron just before it hits a television screen after being accelerated from rest by the 29000V of the picture tube?

Answer 1

The kinetic energy will be equal to the change in electric potential energy
\[K=U\]
\[K=qV\]
And since the charge on an electron is given as e, its energy can be expressed as eV
\[K=29000eV\]
\[\frac{1}{2}m_ev^2=29000eV\]
\[\hspace{2 cm} 1eV = 1.602 \times 10^{-19}J\]

solve for v ^_^

Answer 2

Thanks! That's what I was doing; it turns out masteringphysics had the wrong answer stored, but I was able to guess what it was.