A H e + ion is in its first excited state. Its ionization energy is:

Question

A H e + ion is in its first excited state. Its ionization energy is:, 13.60 e V, 48.36 e V, 54.40 e V, 6.04 e V

MARKS App · Accepted Answer

E = - 13.6 z 2 n 2 = - 13.6 2 2 2 2 = - 13.60 e V Total energy of H e + ion is in its first excited state is - 13.60 e V . Hence, its ionization energy will be 13.60 e V

Search any question & find its solution

Looking for more such questions to practice?