A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the 2 nd orbital to 3 rd orbit is (47.2 eV ), find the atomic number of the given atom.

Step-by-step solution

The energy of (n )th orbit is given as (E_n= -R h C z^2 n^2 ) For hydrogen atom, (z=1 ) and (n=1 ), hence (E₁= -R h C 1^2 1^2 =-R h C ) We know that, (R h C=13.6 eV ) ( E₁=-13.6 eV ) For (n=2 ), ( aligned & E₂= -R h C z^2 2^2 = -13.6 z^2 4 & E₂= -13.6 z^2 4 (i) aligned ) Similarly, for (n=3 ), (E₃= -13.6 9 z^2 (ii) ) Given, ( array ll & -13.6 9 z^2- ( -13.6 z^2 4 )=47.2 & 13.6 5 36 z^2=47.2 & z^2=24.98 & z^2 25 & z=5 array )