(2)

Brackett series corresponds to the electronic transition from $n_2=5,6...$ to $n_1=4$ level.

(2)

Wavelength of spectral lines of hydrogen is given by $\frac{1}{\lambda }=R[\frac{{\displaystyle 1}}{{\displaystyle n_1^2}}-\frac{{\displaystyle 1}}{{\displaystyle n_2^2}}]$

For $n_2=3$ to $n_1=2$

$\frac{1}{\lambda }=1.09\times {10}^7[\frac{{\displaystyle 1}}{{\displaystyle {\left(2\right)}^2}}-\frac{1}{{\left(3\right)}^2}]$

$\lambda =656.3$ $\text{nm}$

$\mathrm{A}\to \mathrm{III}$

For $n_2=4$ to $n_1=2$

$\frac{1}{\lambda }=1.09\times {10}^7[\frac{{\displaystyle 1}}{{\displaystyle {\left(2\right)}^2}}-\frac{1}{{\left(4\right)}^2}]\Rightarrow \lambda =486.1$ $\text{nm}$

$\mathrm{B}\to \mathrm{IV}$

For $n_2=5$ to $n_1=2$

$\frac{1}{\lambda }=1.09\times {10}^7[\frac{{\displaystyle 1}}{{\displaystyle {\left(2\right)}^2}}-\frac{1}{{\left(5\right)}^2}]\Rightarrow \lambda =434.1$ $\text{nm}$

$\mathrm{C}\to \mathrm{II}$

For $n_2=6$ to $n_1=2$

$\frac{1}{\lambda }=1.09\times {10}^7[\frac{{\displaystyle 1}}{{\displaystyle {\left(2\right)}^2}}-\frac{1}{{\left(6\right)}^2}]\Rightarrow \lambda =410.2$ $\text{nm}$

$\mathrm{D}\to \mathrm{I}$

Option (2) is correct

(4)

Shortest wavelength in Balmer series when transition of electron from $\infty$ to $n=2$

$\frac{1}{\lambda }=R(\frac{{\displaystyle 1}}{{\displaystyle 2^2}}-\frac{{\displaystyle 1}}{{\displaystyle {\infty }^2}})$

$\frac{1}{\lambda }=\frac{R}{2^2}=\frac{R}{4}$                                             ...(i)

For Brackett series,

$\frac{1}{\lambda \text{'}}=R{\left(1\right)}^2(\frac{{\displaystyle 1}}{{\displaystyle 4^2}}-\frac{{\displaystyle 1}}{{\displaystyle {\infty }^2}})=R\left(\frac{{\displaystyle 1}}{{\displaystyle 4^2}}\right)=\frac{R}{16}$                ...(ii)

Divide (i) by (ii), we get

$\frac{\lambda \text{'}}{\lambda }=\frac{16}{4}=4\Rightarrow \lambda \text{'}=4\lambda$

(3)

When electron jumps from higher orbit to lower orbit, then the wavelength of emitted photon is given by,

      $\frac{1}{\lambda }=R(\frac{{\displaystyle 1}}{{\displaystyle n_f^2}}-\frac{{\displaystyle 1}}{{\displaystyle n_i^2}})$

so,  $\frac{1}{\lambda }=R(\frac{{\displaystyle 1}}{{\displaystyle 2^2}}-\frac{{\displaystyle 1}}{{\displaystyle 3^2}})=\frac{5R}{36} \text{and} \frac{1}{\lambda \text{'}}=R(\frac{{\displaystyle 1}}{{\displaystyle 3^2}}-\frac{{\displaystyle 1}}{{\displaystyle 4^2}})=\frac{7R}{144}$

$\therefore$   $\lambda \text{'}=\frac{144}{7}\times \frac{5\lambda }{36}=\frac{20\lambda }{7}$

(3)

Energy of the photon, $E=\frac{hc}{\lambda }$

  $E=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}} \text{J}$

     $=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-19}}\text{eV}=12.75$ $\text{eV}$

After absorbing a photon of energy 12.75 eV, the electron will reach the third excited state of energy $-0.85$ eV, since the energy difference corresponding to $n=1$ and $n=4$ is 12.75 eV.

$\therefore$   $\text{Number of spectral lines emitted}=\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6$