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Energy of H-atom in the ground state is –13.6 eV, hence energy in the second excited state is
(a) –6.8 eV (b) -3.4 eV (c) –1.51 eV (d) –4.53 eV
from JEE 2002
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1 Answer

 
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Correct answer is option (c) –1.51 eV  

Explaination::

\(2 ^{nd}\ excited \ state \ n=3 E n =− \frac{ 13.06}{n^2} ​ \) -{By formula }

\(E_3=-\frac{13.06}{3^2}=-\frac{13.06}{9}\) 

= -1.51 eV 

 

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