NEUTRONS IN SPACE STATION!!

$Total\quad distance\quad to\quad be\quad covered\quad =\quad 3200\quad Km\quad =\quad 3200\times { 10 }^{ 3 }\\ Speed\quad =\quad 2000\quad m/s\\ Time\quad taken\quad =\quad \frac { distance\quad covered }{ speed } =\frac { 3200\times { 10 }^{ 3 } }{ 2000 } =1600\quad second\\ Now\quad total\quad neutrons\quad =\quad { 10 }^{ 8 }\\ Since\quad half\quad life\quad =\quad 800\quad second,\quad then\quad adter\quad 800\quad second,\quad remaining\\ neutrons\quad =\quad 0.5\times { 10 }^{ 8 }\quad =\quad 5\times { 10 }^{ 7 }\\ After\quad 1600\quad second,\quad remaining\quad neutrons\quad =\quad 0.5\times 5\times { 10 }^{ 7 }\\ =25\times { 10 }^{ 6 }\\ After\quad didving\quad with\quad { 10 }^{ 6 },\quad answer\quad =\quad 25$

The velocity of neutrons are too small to use Lorentz's factor, so we use classical mechanic.

Time for the neutrons to reach the space station is $\displaystyle t = \frac{3200 \times 10^3}{2000} = 1600$ seconds.

Using radioactive decay law:

$\displaystyle N(t)=N_{0}e^{-\frac{\ln 2}{t_{1/2}}t}$

With $t_{1/2} = 800$ seconds and $N_{0}=10^8$ neutrons, we got $N(t) = 25 \times 10^6$ neutrons $\implies$ the answer is $\boxed{25}$ .