Discussion Forum

Consider a hydrogen -like ionised atom with atomic number Z , with  a single electron. In the emission spectrum  of atom, the photon emitted  in the n= 2 to n=1 transition has energy 74.8 eV higher than the photon emitted in the n=3 to n=2 transition.The ionisation energy of the hydrogen atom is 13.6 eV. The value of Z is..........

Answer:

Explanation:

$\triangle E_{2-1}=13.6\times Z^{2}[1-\frac{1}{4}]=13.6 \times Z^{2}[\frac{3}{4}]$

$\triangle E_{3-2}=13.6\times Z^{2}[\frac{1}{4}-\frac{1}{9}]=13.6 \times Z^{2}[\frac{5}{36}]$

 $\therefore \triangle E_{2}=\triangle E_{3-2}+74.8$

$=13.6\times Z^{2}[\frac{3}{4}]=13.6 \times Z^{2}[\frac{5}{36}]$ + 74.8

$=13.6\times Z^{2}[\frac{3}{4}-\frac{5}{36}]=74.8$

$Z^{2}=9$

Z=3