Physics Nuclei

${}_Z{}^{A}X+{}_1{}^{1}H\to {}_2{}^{4}He+{}_8{}^{16}O$

$Z+1=2+8?Z=9X\to F$

$A+1=4+16\Rightarrow A=19$

$\Rightarrow {}_Z{}^{A}X={}_9{}^{19}F$

${}_7{}^{14}N+{}_Z{}^{A}Y\to {}_1{}^{1}H+{}_6{}^{14}C$

$7+Z=1+6?Z=0$

$14+A=1+14?A=1$

$?{}_Z{}^{A}Y={}_0{}^{1}n$

Density of nuclei is order of 10¹⁷ kg m^-3.

$\begin{array}{l}t=\frac{T}{2}\\ \Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T}}\end{array}$

$\Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{1}{2}}=\frac{N_0}{\sqrt[]{2}}$

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

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

$sin\theta =\frac{1}{2}$  

t = 12 cm

$forA,t_{\frac{1}{2}}=4sec$  

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

$m_A=m_{A0}{e}^{-4\times 0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

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

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

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