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As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly in ionized Li atom (Z=3) is
  • a)
    1.51
  • b)
    13.6
  • c)
    40.8
  • d)
    122.4
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
As per Bohr model, the minimum energy (in eV) required to remove an el...
The Bohr model is a simplified model of the atom proposed by Niels Bohr in 1913. It describes the electron orbits around the nucleus as discrete energy levels, and the energy difference between these levels is quantized. According to the Bohr model, the energy levels of an atom are given by the equation:

E = -13.6 Z^2 / n^2

where E is the energy of the level, Z is the atomic number, and n is the principal quantum number.

In the case of a doubly ionized Li atom (Z=3), we need to find the minimum energy required to remove an electron from the ground state. The ground state is when the electron is in the lowest energy level, which corresponds to n=1.

Using the equation above, we can calculate the energy of the ground state:

E = -13.6 * 3^2 / 1^2
= -13.6 * 9 / 1
= -122.4 eV

The negative sign indicates that the energy is lower than the zero reference point. To remove an electron from the ground state, we need to supply energy equal to the magnitude of the energy of the ground state, but with the opposite sign.

So, the minimum energy required to remove an electron from the ground state of a doubly ionized Li atom is 122.4 eV. Therefore, the correct answer is option D.
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Community Answer
As per Bohr model, the minimum energy (in eV) required to remove an el...
E=13.6*n2/z2
=13.6*1/9
=13.6×9 = 122.4 ev
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As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly in ionized Li atom (Z=3) isa)1.51b)13.6c)40.8d)122.4Correct answer is option 'D'. Can you explain this answer?
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