Why are atomic emission spectra called fingerprints?

When white light passes through a cool gas, atoms absorb specific wavelengths. This produces a continuous background with dark lines at those absorbed wavelengths. This is atomic absorption spectra. But when atoms are excited, they release photons at specific wavelengths, producing a dark background

We see a discrete emission spectrum when electrons inside excited atoms or ions in gases fall back from higher energy levels to lower ones. Every transition releases a photon of a specific wavelength. Usually the spectrum appears as sharp, bright lines. Dark gaps separate them. These lines are uniqu

Kindly go through the solution 

Change in surface energy = work done

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}\times 10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63\times 10^{-34}\times 10.2}{12400\times 10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8\times 10^{-27}$           

$v=\frac{6.63\times 10.2\times 10^{-34}}{12400\times 10^{-10}}$

$v=\frac{6.63\times 10.2}{12400\times 1.8}\times 10^3$            

$=\frac{6.63\times 102}{124\times 1.8}=3.02$ ]

= 3 m/s