Physics Nuclei

$forA,t_{\frac{1}{2}}=4sec$  

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

$m_A=m_{A0}{e}^{-4*0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

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

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

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

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

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

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

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

Energy of products = 939 + 940 = 1879
ΔE + E_reactions – E_productor = 1877 – 1879 = -2MeV.
Hence it must capture a γ - ray photon of energy 2MeV.

${}_7^{14}\mathrm{N}$ contains 7 protons and 7 neutrons.

Mass defect,  $\left({}_{}{}_{}\right)\left(\begin{array}{c}\\ \end{array}\right)$ $\mathrm{\Delta }m=\left(7m_H+7m_n\right)-m\left(\begin{array}{c}14\\ 7\end{array}\right)=(7*1.00783u)+(7*1.00867u)-14.00307u$

$=0.11243u$

Binding energy $=0.11243*931.5\mathrm{M}\mathrm{e}\mathrm{V}=104.7\mathrm{M}\mathrm{e}\mathrm{V}$

t? , half life = (log?2)/λ = 3 s
Initially 200 nuclei were there
After 3 half-lives, 9 seconds,
No. of nuclei = (1/2³) (200) = 25
Therefore, the number of nuclei that decayed = 200 – 25 = 175

N = [n + n/10 + n/100 + .]
= n/ (1 - 1/10) = 10/9

$\frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{1}{16}={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{T}}_{1/2}}}$

$\mathrm{t}=4{\mathrm{T}}_{1/2}=20*4=80\text{minutes}$

Kindly consider the following Image 

 

$\begin{array}{rr}& \mathrm{R}={\mathrm{R}}_0(\mathrm{A}{)}^{1/3}\\ & \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}={\left(\frac{125}{64}\right)}^{1/3}=\frac{5}{4}\end{array}$