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  • Need explanation for: The ionization energy of the ionized sodium atom Na+10 is

 The ionization energy of the ionized sodium atom Na+10 is

  • Option 1)

    13.6 eV

     

  • Option 2)

    13.6 x 11 eV

     

  • Option 3)

     (13.6/11) eV

     

  • Option 4)

    13.6 x (112) eV

     

 

Answers (1)

best_answer

 

Energy emitted due to transition of electron -

\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )

\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )

- wherein

R= R hydberg\: constant

n_{i}= initial state \\n_{f}= final \: state

 

 E=E_0.{z^2}(\frac{1}{m^2}-\frac{1}{n^2})

For ionization m=1, n\rightarrow \infty

\therefore E=E_0z^2 =13.6 \times (11)^2 0v


Option 1)

13.6 eV

 

Incorrect

Option 2)

13.6 x 11 eV

 

Incorrect

Option 3)

 (13.6/11) eV

 

Incorrect

Option 4)

13.6 x (112) eV

 

Correct

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Plabita

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