Question

if E1, E2, E3,are respectively the energies of 1st lines of lyman, balmer, paschen, brackett series for hydrogen atom then correct order of these energies is ​

Answer 1

the correct order is E3 <e2<E1

explanation is that lesser energy is released when the electron jumps to less lower orbital that is its N1 increases along lymen.balmee.paschen.bracket.pfund.

Answer 2

Answer:

Energy of first line of Lyman, Balmer, Paschen, Brackett series of hydrogen atom  will be in order: E₁>E₂>E₃>E₄

Therefore option (c) is correct.

Explanation:

When the energy supplied, the electron absorbs a de finite amount of energy and jumps to higher energy levels. The electron in such is said to be excited, The excited state is unstable  and the electrons come back to lower energy level. It gives the radiation of difference in energies of between two levels:

 $\bar{\nu}=\frac{1}{\lambda} =R_{H}[\frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}} ]$                                              ....................(1)

Where $R_{H}$ is Rydberg constant whose value equal to 1.090 × 10⁷m⁻¹.

• Lyman series: The first line will be from $n_{1}=1 \; to\; n_{2}=2$

       Energy of first line will be:  

           $\bar{\nu}=\frac{1}{\lambda} =R_{H}[\frac{1}{1^{2}} -\frac{1}{2^{2}} ]$

                    $=R_{H}[\frac{1}{1} -\frac{1}{4} ]$

                    $=R_{H}[\frac{4-3}{4}]$

                    $=\frac{3}{4} R_{H}$

                    $=0.75R_{H}$

• Balmer series: The first line will be from $n_{1}=2 \; to\; n_{2}=3$

        Energy of first line will be:  

           $\bar{\nu}=\frac{1}{\lambda} =R_{H}[\frac{1}{2^{2}} -\frac{1}{3^{2}} ]$

                    $=R_{H}[\frac{1}{4} -\frac{1}{9} ]$

                    $=R_{H}[\frac{9-4}{36}]$

                    $=\frac{5}{36} R_{H}$                    

                    $=0.138R_{H}$    

• Paschen series: The first line will be from $n_{1}=3 \; to\; n_{2}=4$

       Energy of first line will be:  

           $\bar{\nu}=\frac{1}{\lambda} =R_{H}[\frac{1}{3^{2}} -\frac{1}{4^{2}} ]$

                    $=R_{H}[\frac{1}{9} -\frac{1}{16} ]$

                    $=R_{H}[\frac{16-9}{144}]$

                    $=\frac{7}{144} R_{H}$                    

                            $=0.0486R_{H}$    

• Brackett series: The first line will be from $n_{1}=4 \; to\; n_{2}=5$

        Energy of first line will be:  

           $\bar{\nu}=\frac{1}{\lambda} =R_{H}[\frac{1}{4^{2}} -\frac{1}{5^{2}} ]$

                    $=R_{H}[\frac{1}{16} -\frac{1}{25} ]$

                    $=R_{H}[\frac{25-16}{400}]$

                    $=\frac{9}{400} R_{H}$                    

                    $=0.0225R_{H}$  

Therefore the energy of first line will be in the order:

Lyman> Balmer> Paschen>Brackett series

So, E₁>E₂>E₃>E₄.