Physics Nuclei

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].

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

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$         

Time required = 4 * $t_{\frac{1}{2}}=20yrs$  

X + p ® Y + b

$Q=K_Y+K_b-K_P$  

⇒ $Q+K_P=K_Y+K_b$  

$\Rightarrow Q+K_P>0$  

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$  

Time required = 4 *  $t_{\frac{1}{2}}=20yrs$

$d=4\sqrt[]{3}$ cm (Lateral shift)

By Snell's law

${\mu }_{air}sin60°={\mu }_{g}sin\theta$

θ = 30

$\Rightarrow 1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}sin\theta$  

$sin\theta =\frac{1}{2}$  

t = 12 cm

$forA,t_{\frac{1}{2}}=4sec$  

$=m{A}_0{e}^{-}\frac{0.693}{4}*16$

$m_A=m_{A0}{e}^{-4*0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

$m_B=m_{B_0}{e}^{-2*0.693}$               -(2)

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

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

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

$\therefore \frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt[]{I_{S_4}}+\sqrt[]{I_{S_3}}\right)}^2}{{\left(\sqrt[]{I_{S_4}}-\sqrt[]{I_{S_3}}\right)}^2}={\left(\frac{2+1}{2-1}\right)}^2=\frac{9}{1}$

$Q_{\left(i\right)}=\left(M_{Br}-M_{Kr}\right)c^2>0$

$\begin{array}{l}{Q}_{\left(ii\right)}=\left(M_{Br}-M_{Se}-2M_e\right)c^2>0\\ Q_{\left(ii\right)}=\left(M_{Br}-M_{se}\right)c^2>0\end{array}$