Question

Using Bohr's Atomic Model, match the following columns.

$($Symbols $n$ and $Z$ have their usual meanings$)$

Column I	Column II
$\left(A\right)$ Due to revolving electron, the magnetic field produced at its centre is proportional to	$\left(p\right)n^{-5}$
$\left(B\right)$ Magnetic moment of a revolving electron is proportional to	$\left(q\right)n$
$\left(C\right)$ De-Broglie wavelength of a revolving electron is proportional to	$\left(r\right)Z^3$
$\left(D\right)$ Areal velocity of a revolving electron about the nucleus is proportional to	$\left(s\right)$ Independent of $Z$
	$\left(t\right)Z^{-1}$

• A. $A\to p,sB\to qC\to q,rD\to p$
• B. $A\to p,rB\to q,sC\to q,tD\to q,s$
• C. $A\to sB\to t,qC\to p,rD\to p,s$
• D. $A\to p,rB\to q,rC\to pD\to q,t$

Answer 1

The correct option is B $A\to p,rB\to q,sC\to q,tD\to q,s$

$\left(A\right)$.
The revolving electron is equivalent to a current carrying circular loop.


The magnetic field at the centre of the loop is,

$B=\frac{{\mu }_{0}i}{2r}$

$\because i=\frac{q}{T}=\frac{e}{\left(\frac{2\pi r}{v}\right)}$

$\Rightarrow B=\frac{{\mu }_{0}ev}{\left(2\pi r\right)\left(2r\right)}=\frac{{\mu }_{0}ev}{4\pi r^2}$

$\Rightarrow B\propto \frac{v}{r^2}$

$\Rightarrow B\propto \frac{\left(Z/n\right)}{(n^2/Z{)}^2}$

$\Rightarrow B\propto \frac{Z^3}{n^5}$

Hence, $B\propto Z^3$ and $B\propto n^{-5}$

$\therefore A\to p,r$

$\left(B\right)$.
The magnetic moment of a revolving electron is,

$M=iA$

$\Rightarrow M=\pi r^{2}i=\frac{\pi r^2\left(ev\right)}{2\pi r}$

$\Rightarrow M\propto rv$

$\Rightarrow M\propto \left(\frac{n^2}{Z}\right)\left(\frac{Z}{n}\right)$

$\Rightarrow M\propto n$

$\therefore B\to q,s$

$\left(C\right)$.
The De-Broglie wavelength of revolving electron,

$\lambda =\frac{h}{mv}$

$\Rightarrow \lambda \propto \frac{1}{v}$

$\Rightarrow \lambda \propto \left(\frac{n}{Z}\right)$

Hence, $\lambda \propto n$ and $\lambda \propto Z^{-1}$

$\therefore C\to q,t$

$\left(D\right)$.
Now, Areal velocity is given by,

$v_A=\frac{L}{2m}$

$\Rightarrow v_A=\frac{nh/2\pi }{2m}$

$\Rightarrow v_A\propto n$

$\Rightarrow D\to q,s$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(B\right)$ is correct.