Question

The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10−18 JJ of kinetic energy in this distance?

Answer 1

Answer:

$a=1.1*10^{18}\frac{m}{s^2}$

Explanation:

We use the following kinematic formula to calculate the acceleration:

$v_f^2=v_0^2+2ax$

The kinetic energy is defined as:

$\Delta K=\frac{m(v_f^2-v_0^2)}{2}\\v_f^2-v_0^2=\frac{2\Delta K}{m}$

Replacing this in the acceleration formula and solving for a:

$\frac{2\Delta K}{m}=2ax\\a=\frac{\Delta K}{mx}\\a=\frac{2*10^{-18}J}{(9.1*10^{-31}kg)(2*10^{-6}m)}\\a=1.1*10^{18}\frac{m}{s^2}$

Answer 2

The solution is in the attachment