Physics Nuclei

${\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.

The number of neutrons in nucleus has an important role in determining the stability of nucleus. Stability of nucleus is affected by the balance between number of protons and neutrons and the number of nucleons. In lighter elements, the most stable nuclei has roughly equal number of protons and neut

Most elements exist as a mixture of isotopes in nature. Isotopes are atoms of same elements that have same number of protons but they have different numbers of neutrons. Since the number of neutrons may vary, different isotopes of same element have different atomic masses.

The atomic mass of an eleme

${\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.

$A\underset{}{\overset{{\lambda }_A}{\to }}B\underset{}{\overset{{\lambda }_B}{\to }}C$

Amount of B,

$N_B=\frac{{\lambda }_A{N}_{A_0}}{{\lambda }_B-{\lambda }_A}\left[e^{-{\lambda }_{A}t}-e^{-{\lambda }_{B}t}\right]$

where K = $\frac{{\lambda }_A}{{\lambda }_B-{\lambda }_A}{N}_{A_0}$

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{t{|}_{t_1|2}}$

$\Rightarrow N=10^{10}\times {\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{10^{10}}{\sqrt[]{2}}?7\times 10^9$

$\frac{d{N}_B}{dt}=\lambda N_A-\lambda N_B=2\lambda N_B\left(0\right).e^{-\lambda t}-\lambda N_B$                

$\Rightarrow \frac{d{N}_B}{dt}+\lambda N_B=2\lambda N_B\left(0\right)e^{-\lambda t}$              

Solving it, $N_B\left(t\right)=N_B\left(0\right)\left(1+2\lambda t\right)e^{-\lambda t}$  

N_B (t) is maximum when,

$\frac{d{N}_B}{dt}=0\Rightarrow t=\frac{1}{2\lambda }$              

We know, R = R_oA₃₌

_{$\Rightarrow \frac{R}{R_O}=A^3$}

Taking log both side

$lo{g}_e\left(\frac{R}{R_O}\right)=3logA$