Question

The energy of an electron in the second excited state of helium+ ion is

Answer 1

Answer:

-6.04eV

Explanation:

we have to use this formula,

E = -13.6 Z2/n2

-13.6eV is for ground state energy of helium+ ion

Z is atomic number, which is 2 for helium.

n is excited state, n is 3 for 2nd excited state.

put those values in faormula,

E = -13.6 (4/9)

E = -6.04eV