Bohr's model of atom - class-X

Use the formula,
$ 1/\lambda = R \times \left ( 1/n _1^2 - 1/n _2^2 \right) $.
Therefore,
$ \Delta E = hc/\lambda = hcR \times \left ( 1/n _1^2 - 1/n _2^2 \right) $.
for Lyman series lowest energy transition is from n$=1 $ to n$=2$.

Use the formula,
Wavenumber = $ R \times \left ( 1/n _1^2 - 1/n _2^2 \right ) $.
For third line Balmer series $ n _1=2$ and $n _2=5 $.
For second line Paschen series $ n _1=3$ and $n _2=5 $
Therefore the ratio is $ \left (1/4 - 1/25 \right) / \left (1/9 - 1/25 \right) $ which equals 
$ 21/16 \times 9/4 $. 

The $\beta$ line of any series means the second line of that series. Similarly the $\alpha$ line implies the first line and the $\gamma$ line implies the third line of the series.

Use the formula ,
wavenumber =R×(1/n21−1/n22)=R×(1/n12−1/n22).

No. of electrons in$Li^{+}$ & $He = 2$. Therefore the spectrum will be same for both of them with same no. of electrons.

$\dfrac {1}{\lambda}=R _HZ^2[\dfrac {1}{n _1^2}-\dfrac {1}{n _2^2}]$
For Balmer Series' first emission line,
$\dfrac {1}{\lambda}=R _H\times 1[\dfrac {1}{2^2}-\dfrac {1}{3^2}]=\dfrac {R _H5}{36}cm^{-1}$

                              Lyman series                           Balmer series
Limiting line   :         $n=\infty :to:n=1$               $n=\infty :to:n=2$

                                                    $E _l > E _B$                           [limiting case]
                                                    $\lambda _l < \lambda _B$
Rydberg constant represents the limiting value of the highest wavenumber of any photon that can be emitted from an atom.
                                                    $\frac{1}{\lambda _l }=Z^2R        [\lambda _l =limiting :case]$

From Bohr model, we know that
in a transition,
$\displaystyle\frac{1}{\lambda}=R _H \left[ \frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right ]$ where, $R _H = 109700cm^-$
For Balmer series, electron gets deexcited from 3rd or upper level to second level so $n _1 =2, n _2 >2$.
Also we know that,
$E=hc/\lambda$ so
$\displaystyle v=\frac{4c(in msec^{-1})}{364.56\times 10^{-9}}\left[\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$; $c$ is speed of light.

In all the three transitions $\displaystyle n=4\rightarrow n=2$, $\displaystyle n=3\rightarrow n=1$ and $\displaystyle n=4\rightarrow n=1$, one quantum of energy is emitted. Whenever an electron jumps from higher energy level to lower energy level, a quantum of energy is emitted.

Balmer $6^{th}$ line (8---2)
Paschen $9^{th}$ line (12---3)
For $\displaystyle He^{+}\frac{1}{\lambda }=R.2^{2}\left ( \frac{1}{2^{2}}-\frac{1}{8^{2}} \right )$

Energy of electron in $n^{th}$ shell:

In the emission spectra of hydrogen atom, Paschen series is the one where the transition from higher energy states to third energy state takes place.