JEE MAIN - Chemistry (2015 (Offline) - No. 22)
Which of the following is the energy of a possible excited state of hydrogen?
–6.8 eV
–3.4 eV
+6.8 eV
+13.6 eV
Paskaidrojums
Energy(En) = $$ - {{13.6{Z^2}} \over {{n^2}}}eV$$
where n = 1, 2, 3, 4, ......
For hydrogen Z = 1 and excited state starts from n = 2
So putting n = 2
E2 = $$ - {{13.6} \over {{2^2}}}eV$$ = -3.4 eV
where n = 1, 2, 3, 4, ......
For hydrogen Z = 1 and excited state starts from n = 2
So putting n = 2
E2 = $$ - {{13.6} \over {{2^2}}}eV$$ = -3.4 eV