Answer
25 collisions.
Work Step by Step
With each collision, one-half of the kinetic energy is lost. In that case, we can find the energy after N collisions.
$$E_N=E_0\left(\frac{1}{2} \right)^N$$
Use the given information.
$$0.040eV=(1.0\times10^6 eV)\left(\frac{1}{2} \right)^N$$
$$\frac{0.040eV}{1.0\times10^6 eV}=\left(\frac{1}{2} \right)^N$$
$$N=\frac{ln\frac{0.040eV }{1.0\times10^6 eV }}{ln\frac{1}{2}}=24.58$$
We round up to 25 colliisons.
