Question

Match the following column

Answer 1

$\text{Part A}$

We know,

$r_n=r_0\frac{n^2}{Z}$ $r_0=0.53pm$

It is clear that $r_n\propto Z^{-1}$

Hence $\text{A}\to 2$

$\text{Part B}$

Current by a revolution of a charged particle=$q\omega =\frac{q\times v_n}{2\pi r_n}$

Current $\propto v_n{r}_n^{-1}$

And velocity of electron in $n^{th}$ orbit$=v_0\frac{Z}{n}$

Hence current $\propto \frac{Z^1}{Z^{-1}}\propto Z^2$

Hence $\text{B}\to 3$

$\text{Part C}$

Magnetic field at centre of current loop$\left(B\right)=\frac{{\mu }_0{I}_n}{4\pi r_n}$

$B\propto \frac{I_n}{r_n}$

$B\propto \frac{Z^2}{Z^{-1}}\propto Z^3$

Hence $\text{C}\to 4$

$\text{Part D}$

$v_n=v_0\frac{Z}{n}$

$v_n\propto Z$

Hence $\text{D}\to 1$