Physics Nuclei

$A=Ao{e}^{-\lambda t}$

$\Rightarrow 2250=4250e^{-\lambda t}$

$\Rightarrow \lambda *10*0.434=0.274$

$\Rightarrow \lambda =\frac{0.274}{10*0.434}=0.63mi{n}^{-1}$

$A\to 286$

$Z=81$

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.

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

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

As we know that equivalent half life of radioactive material decays by simultaneous emissions, is given as

$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{700}+\frac{1}{1400}=\frac{3}{1400}\Rightarrow T=\frac{1400}{3}years$

$\Rightarrow t=\frac{\mathcal{l}n\left(3\right)}{\lambda }=\frac{\mathcal{l}n\left(3\right)}{\mathcal{l}n\left(2\right)}T=\frac{1.1}{0.693}*\frac{1400}{3}\approx 740.74years$

Using conservation linear momentum, we can write

$Mv=\frac{h}{\lambda }=\frac{h\nu }{c}\Rightarrow v=\frac{h\nu }{Mc}$                                                           

Loss of internal energy = Gain in Kinetic energy + energy of gamma ray

= $\frac{1}{2}M{v}^2+h\nu =\frac{1}{2}M{\left(\frac{h\nu }{Mc}\right)}^2+h\nu =h\nu \left[1+\frac{h\nu }{2M{c}^2}\right]$  

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

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

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

${\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_x={\left(t_{mean}\right)}_y$
$\frac{\mathcal{l}n2}{{\lambda }_x}=\frac{1}{{\lambda }_y}$

${\lambda }_x=\left(\mathcal{l}n2\right){\lambda }_y$              

Given N_x = N_y = N₀

So activity A = $\lambda N$

$As{\lambda }_x<{\lambda }_y\Rightarrow A_x<A_y$           

So y will decay faster than x.