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AIIMS 26 May 2019 Morning Question Paper with solutions for online practice
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© examsnet.com
Question : 3
Total: 146
15 eV is given to e ñ in 4 th orbit then find it's final energy when it comes out of H-atom
14.15 eV
13.6 eV
12.08 eV
15.85
Validate
Solution:
Energy of 4th orbit of H-atom =
−
13.6
×
1
16
=
−
0.85
e
V
So energy release = total energy ñ ionization energy of 4
=
15
−
0.85
=
14.15
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