# Nuclear reactions. Star Wars

#### Nuclear Reactions

Please find the value of Q that is released in the nuclear reactions

$^{40}Ca rightarrow e^+ + 40K + nu$

$^{98}Ru rightarrow 4He + 94Mo$

$^{44}Nd rightarrow 4He + 140Ce$ $1 amu =931.5 MeV$ ($E =m*c^2$)

The atomic masses are taken from NIST Atomic Masses

1.

$40Ca rightarrow e^+ + 40K + nu + Q$

$M(40Ca) =39.96259 amu$

$M(40K) =39.96399 amu$

$M(e^+) =9.45*10^{-4} amu$

$M(neutrino) =0 amu$ (only a few eV)

$Q = 931.5*(39.96259 – 39.96399 -9.45*10^{-4}) = -931.5*0.002345 = -2.184 MeV$

2.

$^{98}Ru rightarrow 4He + 94Mo+ Q$

$M(^{98}Ru) =97.905287 amu$

$M(^4He) =4.002603 amu$

$M(^{94}Mo) =93.90509 amu$

$Q =931.5*(97.905287 – 4.002603 – 93.90509) =-931.5*0.002406= -2.2412 MeV$

3.

$^{144}Nd rightarrow 4He + 140Ce + Q$

$M(^{144}Nd) =143.91009 amu$

$M(^4He) =4.002603 amu$

$M(^{140}Ce) =139.905439 amu$

$Q = 931.5*(143.91009 – 4.002603 – 139.905439) =931.5*0.002048 =1.907712 MeV$

Observation: the first two reactions have to be “activated”, (the resulting energy is negative). The 3rd reaction occurs naturally (the resulting energy is positive).

#### Star Wars

The anti missile defense laser used in the “star wars” project focuses a 3.0 $mu m$ infrared light on to a missile situated at 2200 km ways, in a spot of 50 cm diameter. Please find the diameter of the mirror necessary for this size of the spot, taking in consideration the diffraction. The rays that arrive at the focus, will come from different parts of the mirror. There will be a path difference between rays reflected by the center of the mirror and rays reflected by the margins of the mirror. Because of this path difference on the focus there will be maxima and minima of diffraction.

The diameter of the first maxima of diffraction (on the focus) is $d =50 cm$. Focus distance is $F =2200 Km$.

The relation between these is given by (D is the mirror diameter)

$d =1.22*lambda *(F/D)$

$D = 1.22*lambda*F/d = 1.22*3*10^-6*2200*10^3/0.5 =16.104 m$