The excitation energy of first excited state of a hydrogen like atom is $40.8 e V$ . Find the energy needed to remove the electron to form the ion.

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Solution

The excitation of energy can be calculated as :
$△ E = E_{2} - E_{1} = 13.6 Z^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) 40.4 = 13.6 \times \frac{3}{4} \times Z^{2} Z = 2$
The energy required to remove from ground state is:
$\frac{13.6 Z^{2}}{1^{2}} = 13.6 Z^{2}$
= $54.4 e V$ is the answer.

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The excitation energy of first excited state of a hydrogen like atom is 40.8eV. Find the energy needed to remove the electron to form the ion.

The excitation of energy can be calculated as :△E=E2−E1=13.6Z2(112−122)40.4=13.6×34×Z2Z=2The energy required to remove from ground state is:13.6Z212=13.6Z2= 54.4eV is the answer.

The excitation energy of first excited state of a hydrogen like atom is $40.8 e V$ . Find the energy needed to remove the electron to form the ion.

The excitation of energy can be calculated as :
$△ E = E_{2} - E_{1} = 13.6 Z^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) 40.4 = 13.6 \times \frac{3}{4} \times Z^{2} Z = 2$
The energy required to remove from ground state is:
$\frac{13.6 Z^{2}}{1^{2}} = 13.6 Z^{2}$
= $54.4 e V$ is the answer.