Nuclei

 = 1/λ = 30*10? ³ sec.
C = 200µF, R
Q/A = constant [Given]
A = A? e? & Q = Q? e? /?
Q/A = (Q? /A? ) e^ (-t/RC + λt) = constant
For this to be constant, the exponent must be zero.
-t/RC + λt = 0
1/RC = λ
R = 1/ (λC) = /C = 30*10? ³/200*10? = 150 Ω

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_1}{2^6}=\frac{N_2}{2^3}\Rightarrow \frac{N_1}{N_2}=8$

Apply energy conservation,

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}*m*\frac{GM}{3R}$

$\Rightarrow K_i=-\frac{1}{6}\frac{GMm}{R}+\frac{GMm}{R}$

$K_i=\frac{5}{6}\frac{GMm}{R}$

$A\to 286$

$Z=81$

As we all know that like charges repel each other. This means that there must be a repulsive electromagnetic force between positively charged protons. However, the nuclear force is said to be 100 times stronger than electromagnetic force which makes it strong enough to hold protons together within a nucleus. This nuclear force is much stronger than coulomb force at short ranges.

Using conservation of momentum

=> 180 V₁ = 4V₂

=> V₂ = 45 V₁

  $Q=K{E}_{180}+K{E}_{\alpha }$              

$Q=\frac{1}{2}180V_1^2+\frac{1}{2}4V_2^2$        

= $\frac{1}{2}*180*{\left(\frac{V_2}{45}\right)}^2+\frac{1}{2}4V_2^2$

$K{E}_{\alpha }=\frac{5.5MeV}{\left(\frac{46}{45}\right)}=5.38MeV$

$A=-\frac{dN}{dt}=\lambda N=\lambda N_0{e}^{-\lambda t}\Rightarrow \lambda \frac{N_0}{16}=\lambda N_0{e}^{-80\lambda }\Rightarrow \frac{1}{16}=e^{-80\lambda }\Rightarrow \frac{1}{16}=e^{-80\lambda }$

$\Rightarrow 80\lambda =\mathcal{l}n\left(16\right)=4\mathcal{l}n\left(2\right)\Rightarrow t_{\frac{1}{2}}=\frac{\mathcal{l}n\left(2\right)}{\lambda }=\frac{80}{4}=20days$

$R=-\frac{dN}{dt}=\lambda N=\lambda N_0{e}^{-\lambda t}\Rightarrow \mathcal{l}n\left(R\right)=\mathcal{l}n\left(\lambda N_0\right)-\lambda t,so$

$\lambda =tan\theta =\frac{6}{40}=\frac{3}{20}{s}^{-1}\Rightarrow$ Slope of graph

$\Rightarrow t_{\frac{1}{2}}=\frac{0.693}{\lambda }=\frac{0.693*20}{3}=4.62s$