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The ionization energy of hydrogen atom is -13.6eV. The enrgy required to excite the electron in a hydrogen atom from the ground state to the first excited state is ( Avogadro’s constant = 6.022 * 10^23) ​A) 1.69 * 10^(-20) J ​B) 1.69 * 10^(-23) J ​C) 1.69 * 10^(23) J ​D) 1.69 * 10^(25) J

The ionization energy of hydrogen atom is -13.6eV. The enrgy required to excite the electron in a hydrogen atom from the ground state to the first excited state is ( Avogadro’s constant = 6.022 * 10^23)
 
​A) 1.69 * 10^(-20) J
​B) 1.69 * 10^(-23) J
​C) 1.69 * 10^(23) J
​D) 1.69 * 10^(25) J
 

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Grade:12th pass

1 Answers

Arun
25750 Points
3 years ago
The different energy levels of a Hydrogen atom are given by the equation:
 
E = - E0/n2
 
where E0 = -1.312 KJ/mol  and n = 1,2,3… and so on.
 
For ground state n = 1
 
E1=E0/(1)2
 
So that the ground state has energy E1= -1.312 KJ/mol  
 
For first excited state n = 2  
 
So E2=-1.312 KJ/mol /4= -0.328 KJ/mole
 
Energy difference between two states:
 
        ΔE= E2-E1= (-0.328)-(-1.312)
 
                          = 0.984 KJ/mol
 
0.984 KJ/mol energy is required to excite an electron in hydrogen atom from ground to excited state
 
 

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