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QAIPMT-2002-Ch-2766: NEET 2002 Chemistry - Class 11 AIPMT-2002-Ch-2766 Solution: In hydrogen atom, energy of first excited state is -3.4 eV.…
In hydrogen atom, energy of first excited state is -3.4 eV. Then find out K.E. of same orbit of hydrogen atom
Multiple Choice Options
- A +3.4 eV
- B +6.8 eV
- C eV
- D +13.6 eV
Step-by-Step Verified Solution
K.E = 1/2 mv 2
= ( � e 2 n h ) 2 _ 2 m [ _ v = 2 � e 2 n h ]
_ Total energy = E n = _ 2 � 2 m e 4 n 2 h 2
= - ( � e 2 n h ) 2 _ 2m = -K.E
_ K.E = - E n
Energy of first excited state is -3.4 eV
_ Kinetic energy of the same orbit (n = 2) will be +3.4 ev