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Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr's model, will be about
- a)53 pm
- b)27 pm
- c)18 pm
- d)13 pm
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its gr...
On the basis of Bohr's modaL, 
For Li++ ion, Z = 3, n = 1 for ground state,
Given, 90 = 53pm
∴
For Li++ ion, Z = 3, n = 1 for ground state,
Given, 90 = 53pm
∴
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Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its gr...
Bohr Model of the Atom
The Bohr model of the atom was proposed by Niels Bohr in 1913. It was based on the idea that electrons orbit the nucleus in specific energy levels.
Formula for Bohr Radius
The Bohr radius (a0) is the distance between the nucleus and the electron in the ground state of hydrogen. It is given by the formula:
a0 = (4πε0h^2)/(me^2)
where ε0 is the permittivity of free space, h is Planck's constant, me is the mass of the electron, and e is the charge of the electron.
Radius of Li+ Ion
When an atom loses one or more electrons, it becomes a positively charged ion. The Li+ ion has lost one electron, so its electron configuration is 1s^2.
According to the Bohr model, the electron in the ground state of Li+ ion will be in the n=1 energy level. Therefore, the radius of the Li+ ion can be calculated using the same formula as for hydrogen, but with the mass of the nucleus (mn) instead of the mass of the electron:
r = (4πε0h^2)/(mnZ^2e^2)
where Z is the atomic number, which is the number of protons in the nucleus.
Substituting the values for Li+ ion, we get:
r = (4πε0h^2)/(6.941×1^2e^2)
= 18.1 pm
Therefore, the radius of Li+ ion in its ground state, according to the Bohr model, is approximately 18 pm. The closest answer option is C, which is 18 pm.
The Bohr model of the atom was proposed by Niels Bohr in 1913. It was based on the idea that electrons orbit the nucleus in specific energy levels.
Formula for Bohr Radius
The Bohr radius (a0) is the distance between the nucleus and the electron in the ground state of hydrogen. It is given by the formula:
a0 = (4πε0h^2)/(me^2)
where ε0 is the permittivity of free space, h is Planck's constant, me is the mass of the electron, and e is the charge of the electron.
Radius of Li+ Ion
When an atom loses one or more electrons, it becomes a positively charged ion. The Li+ ion has lost one electron, so its electron configuration is 1s^2.
According to the Bohr model, the electron in the ground state of Li+ ion will be in the n=1 energy level. Therefore, the radius of the Li+ ion can be calculated using the same formula as for hydrogen, but with the mass of the nucleus (mn) instead of the mass of the electron:
r = (4πε0h^2)/(mnZ^2e^2)
where Z is the atomic number, which is the number of protons in the nucleus.
Substituting the values for Li+ ion, we get:
r = (4πε0h^2)/(6.941×1^2e^2)
= 18.1 pm
Therefore, the radius of Li+ ion in its ground state, according to the Bohr model, is approximately 18 pm. The closest answer option is C, which is 18 pm.
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Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohrs model, will be abouta)53 pmb)27 pmc)18 pmd)13 pmCorrect answer is option 'C'. Can you explain this answer?
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