Match the following. (1) Energy of ground state of H e + (p) 6.04 eV (2) Potential energy of 1 orbit of H atom (q) -27.2 eV (3) Kinetic energy of II excited state of H e + (r) 54.4 eV (4) Ionisation potential of H e + (s) -54.4 eV

Step-by-step solution

Energy of G.S. of H e + = E H × Z 2 = − 13.6 × 4 = − 54.4 eV Ionisation potential of H e + = E ∞ - E 1 = − − 54.4 = 54.4 eV T.E. of H of 1 orbit = − 13.6 eV 2 T.E. = P.E. and K .E = −  T . E P .E . = − 2 × 13.6 = − 27.2 eV T.E. of H of II excited state = - 13.6 9 × 4 of He = − 6.04 eV K .E . = − (T .E .) = − − 6.04 eV = + 6.04 eV