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The ionization energy of 11 times ionized Mg atom is
1. 13. 6 x 102 eV
2. 13.6 × (12)2 eV
3. \(\dfrac{13.6}{(12)^2} \ eV\)
4. 13.6  eV

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Correct Answer - Option 2 : 13.6 × (12)2 eV

Concept: 

The energy that is needed to remove the electron from the atom is called Ionisation Energy. In other words, Ionisation Energy is the energy supplied to an isolated atom to remove its most loosely bound valence shell electron to form a positive ion. The unit of Ionisation energy is electron-Volt(eV). Whereas, Ionization potential is the potential required to move an electron from an orbit to infinity. Its unit is V

Thus, If we need to move an electron revolving in nth orbit to n= \(\infty\) 

Then, Ionization Energy is given by 

I.E.= \(E_\infty - E_n \) 

\(I.E. = 0- E_0\dfrac{Z^2}{n^2}\)

i.e. \(I.E.= 13.6\dfrac{Z^2}{n^2} \)eV 

Ionization potential is equal Ionization energy represented in Volts instead of eV

Calculation:

When Mg is ionized 11 times there is only one electron left in the atom in the n=1 state.

Also, since there is one electron present in the ionized atom so Bohr's Model can be applied. 

We know that,  \(I.E.= 13.6\dfrac{Z^2}{n^2} \)

For 11 times ionized Mg atom, there is only one electron with n=1 

for Magnesium, Z=12

So, ionization energy is given by 

\(\large{(I.E.)_{Mg}}\) =   \(13.6 \times \dfrac{12^{2}}{1^2} \) 

                    \(= 13.6\times (12)^2 \ eV\)

Option(2) is the correct answer

  • The ionization energy \(E_\infty\) that is required to remove the electron from the first Bohr orbit is called the Ionization limit of H- atom.
  • The formula \(E=E_0 \dfrac{Z^2}{n^2}\), is valid only for H- like atom(one electron).

 

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