Question

As per bohr model, the minimum energy required to remove an electron from the ground state of doubly ionized li atom (z = 3) is

Answer 1

Answer:

Explanation:

Energy of electron in the nth state hydrogen like atom = -13.6×z^2÷n^2 ev

Then,

-13.6×3^2÷1^2= -13.6×9×1

=122.4 ev

(. I hope this helps. )

Answer 2

The minimum energy required to remove an electron from the ground state of doubly ionized Li atom is 122.4 eV

Explanation:

Energy of the nth orbit by Bohr was given by:

$E_n=R_H\times \frac{Z^2}{n^2}eV$

where,

E = energy

$R_H$ = Rydberg's Constant

Z = atomic number = 3  (for lithium)

n =  number of orbit  

Putting the values, in above equation, we get

Energy of the first shell(n=1) in hydrogen atom:

Z = 1

$-13.6eV=R_H\times \frac{1^2}{1^2}$

$R_H=-13.6eV$

To find  energy value of electron in the excited state of $Li^{2+}$ is:

$Li:1s^22s^1$

$Li^{2+}:1s^1$

Z = 3 , n= 1

$E_n=-13.6\times \frac{3^2}{1^2}eV=-122.4eV$

Thus the minimum energy required to remove an electron from the ground state of doubly ionized Li atom is 122.4 eV

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