Class 11th

$E_n=-R_H\left(\frac{Z^2}{n^2}\right)J$

For ${}^{+}\left(n=1\right)$ ,

$E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4R_H$

$\therefore R_H=\frac{x}{4}$

For ${}^{3+}\left(n=2\right)$ ,

$E_n=-R_H\left(\frac{z^2}{n^2}\right)J$

$=-\frac{x}{4}*\left(\frac{4*4}{2*2}\right)=-xJ$

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

(A) Isothermal process $\Rightarrow$ Temperature is constant throughout the process

(B) Isochoric process $\Rightarrow$ Volume is constant throughout the process

(C) Isobaric process $\Rightarrow$ Pressure is constant throughout the process

(D) Adiabatic process $\Rightarrow$ No exchange of heat $\left(q\right)$ between system and surrounding

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

Capacitive Reactance $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2*3.14*50*10*10^{-6}}$

$=\frac{1000}{3.14}$

$V_{rms}=210 V$

$i_{rms}=\frac{V_{rms}}{X_C}=\frac{210}{X_C}$

$Peak current =\sqrt[]{2}{i}_{ms}=\sqrt[]{2}*\frac{210}{1000}*3.14=0.932$

$?0.93 A$

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

As $Q_c>K_c$ , so reaction has a tendency to move backward.

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$

$T=m\widehat{?}{\omega }^2$

$T^{\prime }=m:{(2\omega )}^2$

$T^{\prime }=4T$

(Material)                      (Susceptibility ( $\chi$  )

Diamagnetic                 (II) $0>\chi \ge -1$

Ferromagnetic  (III) $\chi ?1$

Paramagnetic                (IV) $0<\chi <\epsilon$

Non-magnetic               (I) $\chi =0$