Physics Ncert Solutions Class 12th

The relation between $\alpha$ and $\beta$  is given as

$\beta =\frac{\alpha }{1-\alpha }$

${\beta }_1=\frac{\frac{20}{21}}{1-\frac{20}{21}}=20$

${\beta }_2=\frac{\frac{100}{101}}{1-\frac{100}{101}}=100$

$\Rightarrow l=\frac{2}{50}=40mA$

Flux through a surface $\mathrm{?}=\stackrel{?}{E}.\stackrel{?}{A}$  

$\mathrm{?}=\left(6\stackrel{?}{i}-4\stackrel{?}{j}+7\stackrel{?}{k}\right).200\stackrel{?}{i}$

$=1200unit$

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

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

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

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

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

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

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

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

In reverse biasing, due to collision of electrons and atom, avalanche breakdown occurs.

$l=\frac{90?30}{4000}=15mA$

$I_1=\frac{30}{5000}=6mA\&l_2=9mA$

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n=\frac{90k{H}_z}{2*5kHz}=9$

Kindly go through the solution

 $y=\overline{\overline{A}+B}=A.\overline{B}$

B_v = B sin 60°

-> $B_v=2.5*10^{-4}*\frac{\sqrt[]{3}}{2}$

$Emf=B_v*v*\mathcal{l}=2.5*10^{-4}*\frac{\sqrt[]{3}}{2}*180*\frac{5}{18}*1=108.25*10^{-3}volts$        

$A=A_0{e}^{-\lambda t_1}$     [Radio active decay law]

$\frac{A}{5}=A_0{e}^{-\lambda \left(t_2-t_1\right)}$

$\Rightarrow Averagelife=\frac{1}{\lambda }=\frac{\left(t_2-t_1\right)}{\mathcal{l}n5}$