In a hydrogen like atom, energy required to excite the electron from its first excited state to second excited state is 7.55 eV . The energy required to remove the electron from its ground state is

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The energy difference between the second excited state ( $n = 3$ ) and the first excited state ( $n = 2$ ) is given as 7.55 eV.

For a hydrogen-like atom, the energy levels are given by

$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$ eV.

Thus, the energy difference is:

$E_{3} - E_{2} = 13.6 Z^{2} (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = 7.55$

$13.6 Z^{2} (\frac{1}{4} - \frac{1}{9}) = 7.55$

$13.6 Z^{2} (\frac{5}{36}) = 7.55$

$Z^{2} = \frac{7.55 \times 36}{13.6 \times 5} = \frac{271.8}{68} = 4.0$

$Z = 2$

The ionization energy (energy to remove an electron from the ground state, $n = 1$ ) is:

$E = 13.6 Z^{2} = 13.6 \times 2^{2} = 13.6 \times 4 = 54.4$ eV.

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