# Nuclear Physics

1. Fill in the missing particles or nuclei

2. The isotope (32/15) P is produced by the reaction

4. Does the reaction  $p + (7/3)Li = (4/2) He + \alpha$ require energy or does it release energy? How much energy?

16. Calculate the energy released in the fission reaction  $n + (235/92) U = (88/38) Sr + (136/54) Xe + 12n$

26. Show that the energy released in the fusion reaction $(2/1)H + (3/1)H = (4/2)He + n$ is 17.59 MeV

1.

a) $n + (137/56)Ba = (138/56)Ba + \gamma$

b) $n + (137/56)Ba = (137/55)Cs +(1/1)H$ (hydrogen),     $(1/1)H$ is the same thing as $p$ (proton)
c) $(2/1)d + (2/1)H = (4/2)He + \gamma$        here $d$ and $(2/1)H$ is the same thing
d) $(4/2)alpha + (197/79)Au = (199/80)Hg + (2/1)d$

2.

$(1/0)n + (32/16)S = (32/15)P + p$

4.

$p + (7/3)Li = (4/2)He + (4/2)\alpha$ ,   here $\alpha$ is the same thing as $(4/2)He$

Atomic masses of isotopes are as follows

$M(p) =1.00728$ amu

$M(7/3)Li =7.016$ amu

$M(\alpha) = 4.0026$ amu

Mass defect is

$\delta(m) = 1.00728+7.016 -2*4,0026 =0.0181$ amu it is positive hence the energy is released from the reaction

energy is

$E = \Delta(m)*c^2 = 0.0181*1.66*10^{-27} *(3*10^8)^2 = 2.7*10^{-12} Joules = 16.9$ MeV

16.

$n + (235/92)U = (88/38)Sr + (136/54)Xe + 12n$

Atomic masses of isotopes are as follows

$M(n) = 1.00866$ amu

$M(235/92)U = 235.044$ amu

$M(88/38)Sr = 87.9056$ amu

$M(136/54)Xe =135.907$ amu

mass defect is

$\delta(m) =1.00866 + 235.044 -87.9056 -135.907 -12*1.00866 =0.136$ amu

energy released is

$E = \Delta(m)*c^2 = 0.136*1.66*10^{-27}*(3*10^8)^2 =8.078*10^|{-12} joules =50.49$ MeV

26.

$(2/1)H + (3/1)H = (4/2)He + n$    or   $D +T = \alpha + n$
Atomic masses are

$M(D) = 2.014$ amu

$M(T) = 3.016$ amu

$M(\alpha) = 4.0026$ amu

$M(n) = 1.00866$ amu

mass defect is

$\Delta(m) = 2.014+3,016-4,0026-1,00866 =0.01874$ amu

energy released is

$E = \Delta(m)*c^2 =0.01874*1.66*10^{-27}*(3*10^8)^2 =2.799*10^{-12} J =17.50 MeV$