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Energy of H-atom in the ground state is -13.6 eV, find energy in the second excited state

Energy of H-atom in the ground state is -13.6 eV, find energy in the second excited state

Grade:11

2 Answers

Vikas TU
14149 Points
4 years ago
Hiii 
The energy in the ground state = -13.6eV 
Z of Helium atom = 2
n = 3 since it is second excited state.
En = Egrund * Z^2/n^2
=> -13.6 * 4/9
=> -6.044 eV 
 
Rajat
213 Points
4 years ago
The formula is
En= K(Z^2/n^2)
Where K= -13.6 eV
Z is the atomic number which 1 here for HYDROGEN atom
and n is 3 for second excited state.
En= -13.6(1/9)
= -1.51111 eV

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