The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from three lo

Question

The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from three lo

The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from three lowest energy orbits of hydrogen atom respectively are
A. 13.6,6.8,8.4 eV
B. 13.6, 10.2, 3.4 eV
C. 13.6, 27.2 40.8 eV
D. 13.6,3.4,1.5 eV

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`E_(T) =- 13.6 (Z^(2))/(n^(2))`
I orbit for (H)
Z=1,n=1
`E_(T)=- 13.6 eV`
II orbit for (H)
Z=1,n=2
`E_(T)=- (13.6)/(4) =- 3.4 eV`
III orbit for (H)
`E_(T) =- (13.6)/(4) =- 1.51 eV`

4 views Answered 2023-02-05 15:29