(1)

As force (F) is constant, so

$F=\frac{m{v}^2}{r}$

We can write,  $F=\frac{m^2{v}^2{r}^2}{m{r}^3}=\frac{L^2}{m{r}^3} \text{(As, }L=mvr\text{)}$

As force is constant, so $L^2\propto m{r}^3$

Using, $L=\frac{nh}{2\pi },$ we get $\frac{n^2{h}^2}{4{\pi }^2}=m{r}^3\Rightarrow n^2\propto r^3 \text{or} r\propto n^{2/3}$

Also, $mvr=\frac{nh}{2\pi }$

As $r\propto n^{2/3},$ so $mv\left(n^{2/3}\right)\propto \frac{nh}{2\pi }$

or   $v\propto n^{1/3}$

(4)

Flux, $\Phi =B\times \pi r^2$

or  $\frac{nh}{e}=B\times \pi r^2 \text{or} r^2=\frac{nh}{Be\pi }$                                  ...(i)

Also, $Bve=\frac{m{v}^2}{r} \text{or} mv=Ber$                                   ...(ii)

Let magnetic moment of electron in circular orbit of an atom be $\mu$.

       $\mu =IA=\frac{e}{T}A=\frac{e\omega }{2\pi }\left(\pi r^2\right)=\frac{evr}{2}$                            ...(iii)

where $m$ is the mass of electron and $L$ is angular momentum.  

From eq. (iii),

         $\mu =\frac{evr}{2}=\frac{e}{2m}L (\because$ $L=mvr)$

or    $\mu =\frac{e}{2m}\times mvr=\frac{e}{2m}\times Be{r}^2 \text{(Using eq (ii))}$

or    $\mu =\frac{e}{2m}\times Be\times \frac{nh}{Be\pi } \text{(Using eq (i))}$

$\therefore$   $\mu =\frac{he}{2\pi m}$

(4)

As, $\Delta E=E_i-E_f$ and $\lambda =\frac{hc}{\Delta E}$

$\Delta E_1=\frac{-5E}{2}-(-4E)=\frac{3E}{2}$

${\lambda }_1=\frac{hc}{\Delta E_1}=\frac{2hc}{3E}$                                          ...(i)

$\Delta E_2=-2E-(-3E)=-2E+3E=E$

${\lambda }_2=\frac{hc}{\Delta E_2}=\frac{hc}{E}$                                            ...(ii)

$\Delta E_3=-2E-(-4E)=-2E+4E=2E$

${\lambda }_3=\frac{hc}{\Delta E_3}=\frac{hc}{2E}$                                           ...(iii)

From equation (i), (ii) and (iii)

${\lambda }_2>{\lambda }_1>{\lambda }_3 \text{and} {\lambda }_2=2{\lambda }_3$

(3)

The energy of $n^{\text{th}}$ state in the hydrogen atom will be $E_n=\frac{-13.6 \text{eV}}{n^2}$

For $n=3;$   $E_3=\frac{-13.6 \text{eV}}{9}=-1.51$ $\text{eV}$

So, ionization energy will be, $-(-1.51)$ $\text{eV}=1.51$ $\text{eV}$

(1)

For angular momentum $\left(L\right)=1.5\frac{h}{\pi }=3\left(\frac{h}{2\pi }\right)$

Hence, $n=3$

$E_3=\frac{-13.6}{n^2}=\frac{-13.6}{9}=-1.51$ $\text{eV}$

(2)

Given, $r_1=5.3\times {10}^{-11}\mathrm{m}$

Radius of $n^{\text{th}}$ orbit of Hydrogen atom is, $r_n=0.53\times \frac{n^2}{Z}$$\mathrm{\AA }$

The radius of the third allowed orbit of hydrogen atom is, $r_3=0.53\times \frac{{{\displaystyle (}{\displaystyle 3}{\displaystyle )}}^2}{1}\mathrm{\AA }$

$r_3=47.7\times {10}^{-11}\text{m}; r_3=4.77\times {10}^{-10}\text{m or }4.77$ $\mathrm{\AA }$

(4)

For $1^{\mathrm{st}}$ excited state, n = 2

For $2^{\mathrm{nd}}$ excited state, n = 3

$\therefore$   $\frac{T_1}{T_2}=\frac{n_2^2}{n_1^2}=\frac{3^2}{2^2}=\frac{9}{4}$

(4)

Bohr’s atomic model is valid for single electron species only. A singly ionised neon contains more than one electron. Hence option (4) is correct.

(4)

Total energy of electron in $n^{\text{th}}$ orbit, $E_n=\frac{-13.6Z^2}{n^2} \text{eV}.$

Kinetic energy of electron in $n^{\text{th}}$ orbit, $\text{K.E.}=\frac{13.6Z^2}{n^2} \text{eV}$

Potential energy of electron in $n^{\text{th}}$ orbit, $\text{P.E.}=\frac{-27.2Z^2}{n^2} \text{eV}$

Thus, total energy of electron, $E_n=-\text{K.E.}=\frac{\text{P.E.}}{2}$

$\therefore$   $\text{K.E.}=3.4 \text{eV}$                                     $\text{[Given }{E}_n=-3.4 \text{eV]}$

        $\text{P.E.}=2\times (-3.4)=-6.8 \text{eV}$

(3)

Given, radius of first Bohr orbit for electron in a hydrogen atom,  $r=0.51$ $\mathrm{\AA }$

and its ground state energy, $E_n=-13.6$ $\text{eV}$

Charge of muon = charge of electron

Mass of muon = 207$\times$(mass of electron)

Therefore, when electron is replaced by muon, then first Bohr radius, $r{\text{'}}_1=\frac{0.51\mathrm{\AA }}{207}=2.56\times {10}^{-13} \text{m}$ and ground state energy, 

$E{\text{'}}_1=-13.6\times 207=-2815.2$ $\text{eV}=-2.815$ $\text{keV}$