Question

the difference between the wave number of 1st lone of balmer series and last line of paschen series for Li2+ ion is

Answer 1

see it and match the answer if correct say!!!

Answer 2

Answer:

The difference between the wave number of 1st lone of balmer series and last line of paschen series is $2.74\times 10^6 m^{-1}$.

Explanation:

The wave-number of the first line in the Balmer series of hydrogen spectrum.

$n_i=3, n_f=2$

$\frac{1}{\lambda }=R\times Z^2[\frac{1}{n_i^{2}}-\frac{1}{n_f^{2}}]$

$\bar\nu=\frac{1}{\lambda }=1.097\times 10^7 m^{-1}\times 3^2[\frac{1}{3^{2}}-\frac{1}{2^{2}}]$

$\bar\nu=1.097\times 10^7 m^{-1}\times \frac{-5}{36}$

The wave-number of the last line in the paschen series of hydrogen spectrum.

$n_i=\infty, n_f=3$

$\frac{1}{\lambda }=R\times Z^2[\frac{1}{n_i^{2}}-\frac{1}{n_f^{2}}]$

$\bar\nu'=\frac{1}{\lambda }=1.097\times 10^7 m^{-1}\times 3^2[\frac{1}{\infty ^{2}}-\frac{1}{3^{2}}]$

$\bar\nu'=1.097\times 10^7 m^{-1}\times \frac{-1}{9}$

The difference between the wave number of 1st lone of balmer series and last line of paschen series is .

$\bar\nu-\bar\nu= 1.097\times 10^7 m^{-1}[9(\frac{-5}{36})-9(\frac{-1}{9})]$

$=1.097\times 10^7 m^{-1}\times \frac{-1}{4}=-2.74\times 10^6 m^{-1}$

Negative sign indicates the energy emission.

The difference between the wave number of 1st lone of balmer series and last line of paschen series is $2.74\times 10^6 m^{-1}$.