Nuclei

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

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

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

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.

$\frac{\mathrm{A}}{{\mathrm{A}}_0}={\left(\frac{1}{2}\right)}^{\mathrm{t}/{\mathrm{T}}_{\mathrm{H}}}={\left(\frac{1}{2}\right)}^{150/100}=\frac{1}{2\sqrt[]{2}}$

$\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{u}}{3}\right)}^2-\frac{1}{2}{\mathrm{m}\mathrm{u}}^2=-{\mathrm{F}}_{\mathrm{R}}\times 24$

$0-\frac{1}{2}{\mathrm{m}\mathrm{u}}^2=-{\mathrm{F}}_{\mathrm{R}}\times \mathrm{d}$

$\frac{\frac{1}{2}{\mathrm{m}\mathrm{u}}^2}{\frac{1}{2}m{u}^2\times \frac{8}{9}}=\frac{\mathrm{d}}{24}$

$\mathrm{d}=24\times \frac{9}{8}=27\mathrm{c}\mathrm{m}$

$\mathrm{}{\lambda }_{\text{electron}}=\frac{h}{\sqrt[]{2mE}}\mathrm{}\mathrm{}\frac{{\lambda }_{\text{electron}}}{{\lambda }_{\text{photon}}}=\frac{1}{c}\sqrt[]{\frac{E}{2m}}$

${\lambda }_{\text{photon}}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{E}}$

A = Activity = λN = (ln (2)/t? /? ) N
N = (1.5×10? ³ / 198) × 6×10²³
A = (0.693 / (2.7×24×3600) × (1.5×10? ³ / 198) × 6×10²³ disintegration/s
A ≈ 1.32 × 10¹³ Bq = (1.32×10¹³ / 3.7×10¹? ) Ci ≈ 357 Ci