Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 year. After 1 year,
(a) All the containers will have 5000 atoms of the material
(b) All the containers will contain the same number of atoms of the material but that number will only be approximately 5000
(c) The containers will in general have different numbers of the atoms of the material but their average will be close to 5000
(d) None of the containers can have more than 5000 atoms

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

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

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

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)

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

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

The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 × 8.5 + 120 × 8.5) - (240 × 7.6) MeV
Q = (2 × 120 × 8.5) - (240 × 7.6) MeV = 2040 - 1824 = 216 MeV.