13.20 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

$r=R_0{\left(192\right)}^{\frac{1}{3}}$

$\frac{r}{2}=R_0{\left(m\right)}^{\frac{1}{3}}$

$m=\frac{192}{8}=24$

Q = [4 *4.0026 – 15.9994] *931.5 MeV

Q = 10.2 MeV

$A=A_0{e}^{-\lambda t_1}$     [Radio active decay law]

$\frac{A}{5}=A_0{e}^{-\lambda \left(t_2-t_1\right)}$

$\Rightarrow Averagelife=\frac{1}{\lambda }=\frac{\left(t_2-t_1\right)}{\mathcal{l}n5}$  

$forA,t_{\frac{1}{2}}=4sec$  

$=m{A}_0{e}^{-}\frac{0.693}{4}\times 16$

$m_A=m_{A0}{e}^{-4\times 0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

$m_B=m_{B_0}{e}^{-2\times 0.693}$               -(2)

$\frac{m_A}{m_B}=\frac{m_{AO}}{m_{BO}}\frac{e^{-4}\times 0.693}{e^{-2}\times 0.693}$

$\frac{m_A}{m_B}=\frac{25}{100}=\frac{x}{100}x=25$