Nuclei

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 neutrons. For instance, Carbon-12 is the most common isotope of carbon with 6 protons and 6 neutrons.

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

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

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,

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

$A=A_0{e}^{-\lambda t}$

$\begin{array}{rr}& 500=700{\mathrm{e}}^{-\lambda \mathrm{t}}\\ & \lambda \mathrm{t}=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & \frac{\mathrm{l}\mathrm{n}?2}{{\mathrm{t}}_{1/2}}*30=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & ?\mathrm{}{\mathrm{t}}_{1/2}=\frac{\mathrm{l}\mathrm{n}?2*30}{\mathrm{l}\mathrm{n}?\frac{7}{5}}\end{array}$

${\mathrm{t}}_{1/2}=61.8\mathrm{m}\mathrm{i}\mathrm{n}\approx 62\mathrm{m}\mathrm{i}\mathrm{n}$

  ${\lambda }_A=25\lambda ,$ ${\lambda }_B=16\lambda$      

At t = 0 Þ N_A = N_B = N₀

after t =  $\frac{1}{a\lambda }:-\frac{N_B}{N_A}=\frac{N_0{e}^{-16\lambda t}}{N_0{e}^{-25\lambda t}}=e^{\left(25\lambda -16\lambda \right)t}$  

->e =  $e^{\left(9\lambda \frac{1}{a\lambda }\right)}$  

$\Rightarrow \frac{9}{a}=1\Rightarrow a=9$      

t_1/2 = 2hr 30 min

Radiation intensity a disintegrations / sec (Activity)

As,   $A=\frac{A_0}{{2^t}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

For safe working, $A=\frac{A_0}{64}\Rightarrow \frac{A_0}{64}=\frac{A_0}{2^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$}\right.}}$  

$\Rightarrow t=6t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=6*2.5hr$                            

= 15

Given A₂ = $\frac{{}_{}}{}$ $\frac{A_1}{2}$

Bernoulli's b/w (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{{\rho }_2}{\rho g}+\frac{v_2^2}{2g}+z_2$

[z₁ = z₂]

$\frac{P_1-P_2}{\rho g}=\frac{v_2^2-v_1^2}{2g}$

[Q = A₁V₁ = A₂V₂]

$\frac{4500}{750}=\frac{{\left(\frac{Q}{A_2}\right)}^2}{2}-{\left(\frac{Q}{A_1}\right)}^2$

$12=Q^2\left[\frac{1}{A_2^2}-\frac{1}{A_1^2}\right]$

$Q=2A_1=2*1.2*10^{-2}$

= 2.4 * 10^-2 m³ /sec

= 24 * 10^-3 m³ /sec

= 24

 $e{V}_0+h{\upsilon }_0=\frac{hC}{\lambda }$

$\Rightarrow 1.6*10^{-19}*0.42+6.63*10^{-34}{\upsilon }_0=\frac{6.63*10^{-34}*3*10^8}{663^0*10^{-10}}$

$\Rightarrow 6.63*10^{-34}{\upsilon }_0=10^{-19}\left(3-1.6*0.42\right)$

$6.63*10^{-34}{\upsilon }_0=2.328*10^{-19}$

${\upsilon }_0=35.11*10^{13}=35$

 $\Delta L$  (change in length) $=\frac{F\mathcal{l}}{AE}=\frac{mg\mathcal{l}}{AE}$

$\Delta \mathcal{l}\propto g$

$\frac{\Delta {\mathcal{l}}_1}{g_1}=\frac{\Delta {\mathcal{l}}_2}{g_2}\Rightarrow \frac{10^{-4}}{10}=\frac{6*10^{-5}}{g_1}$

$g_1=\frac{6*10^{-4}}{10^{-4}}=$ 6 m/sec²