Question

the ionization enthalpy of hydrogen is 1312.0 kj mol-¹ express the value in eV per atom using the relation 1eV =1.602×10^-19j plzzzz plzz helps​

Answer 1

Explanation:

ANSWER

Energy of electron in first orbit =E1=−1312 kJ/mol

Energy of electron in secondorbit =E2=−221312=−328 kJ/mol

Hence, Energy required to excite a hydrogen electron from n=1 to n=2 =E2−E1=−328−(−1312)=984 kJ/mol

Hence B is the correct answer