Nuclei

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

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

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

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

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

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

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,

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

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

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

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

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

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\times 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]

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

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

$Q=2A_1=2\times 1.2\times 10^{-2}$

= 2.4 × 10^-2 m³ /sec

= 24 × 10^-3 m³ /sec

= 24

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

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

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

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

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