The ground state electronic energy in Be3+ ion is

The ground state electronic energy in ${Be}^{3 +} ion$ is

A
$3.49 \times 10^{- 14} J$
B
$3.49 \times 10^{- 15} J$
C
$3.49 \times 10^{- 16} J$
D
$3.49 \times 10^{- 17} J$

Solution:

Ground state energy of

${Be}^{3 +} = Z^{2} E_{H} = (16) (2.18 \times 10^{- 18} J) = 3.49 \times 10^{- 17} J$

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