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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?
9 months 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
9 months ago
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