Atoms

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

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

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

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

  $1.8*10^{-27}$           

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

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

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

= 3 m/s

$-0.85=\frac{-13.6}{n^2}$  

n = 4

Number of transitions = $\frac{4*3}{2}=6$  

Kinetic energy: Potential energy = 1 : –2

Kindly go through the solution

rⁿ? µn²

The spectrum of H-atom is formulated and drawn by Bohr's model. These spectrums are line spectrum. Lyman series lies in UV region, Balmer Partly lies UV and partly in the visible region and Paschen series lies in the infrared region.

The spectrum of H-atom is formulated and drawn by Bohr's model. These spectrums are line spectrum. Lyman series lies in UV region, Balmer Partly lies UV and partly in the visible region and Paschen series lies in the infrared region.