Question

What will be total energy of electronic which is present in the second excited state of $H-$atom?

• A. -3.4 eV
• B. -13.6 eV
• C. -122.4 eV
• D. -1.51 eV

Answer 1

The correct option is D -1.51 eV

Relation between n and E is

$E=\frac{-13.6}{n^2}$

Here it is present in 2nd excited state, So n = 3

$E=\frac{-13.6}{3^2}$

$E=-1.51eV$