Question

Energy of the H- atom in the ground state is -13.6 eV, hence the energy in the second excited state is :

• A. $-6.8eV$
• B. $-3.4eV$
• C. $-1.51eV$
• D. $-4.53eV$

Answer 1

The correct option is B $-3.4eV$

Exitation energy state means atoms jumps from ground state$(n=0)$ to any state

$E=-13.6z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

where $n_1=$final state

$n_2=$initial state

$=-13.6(1{)}^2(\frac{1}{(2{)}^2}-\frac{1}{(0{)}^2})$

$z=1$ for H-atom

$=-3.4eV$.