Energy of Li2+ in first excited state ?
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Answered by
7
hlo dear friend
we have to calculate the energy of first excite
(n,z) =-
Rz^2n^2........1where R =rydberg constant = 2.
2.18×10^-18×91= 19.62×10^-18 joule E = 19.6
hope u understand if not then message me i will explain u how is it.........
we have to calculate the energy of first excite
(n,z) =-
Rz^2n^2........1where R =rydberg constant = 2.
2.18×10^-18×91= 19.62×10^-18 joule E = 19.6
hope u understand if not then message me i will explain u how is it.........
Answered by
0
Answer:
The first excited state of hydrogen is n = 2 (n = 1 is the lowerst-energy, or "ground" state), which from the Rydberg equation has an energy equal to E(n = 2; Z = 1) = -R/4 = -3.4 eV.
Explanation:
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