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Grade: 12th pass
        
THE IONIZATION ENERGY FOR HYDROGEN ATOM IS 13.6 eV THEN REQUIRED ENERGY IN eV TO EXCITE IT FROM THE GROUND STATE TO 1ST EXCITED STATE?
one year ago

Answers : (1)

Apurva Sharma
196 Points
							
E=13.6 Z2/n2 eV/atom
thus for 1st exited state of H, n=2
and for H,Z=1
thus putting the values in the equation
E= 13.6(1)2/22 = 3.4 eV/atom energy
one year ago
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