Question

Instructions:Select the ONE correct answer from the given options.

In hydrogen atom, energy of first excited state is -3.4 \mathrm{eV}−3.4eV. Then, \mathrm{KE}KE of same orbit of hydrogen atom is
 

(1) 
+13.6 \mathrm{eV}+13.6eV

(2) 
+3.4 \mathrm{eV}+3.4eV

(3) 
+6.8 \mathrm{eV}+6.8eV

(4) 

​

Answer 1

+13.6/ mathrm(ev)+13.6ev