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I have a doubt, kindly clarify. - Atoms And Nuclei - JEE Main

Energy required for the electron excitation in L^{++} from the first to the third Bohr orbit is

  • Option 1)

    12.1\: eV

  • Option 2)

    36.3\: eV

  • Option 3)

    108.8\: eV

  • Option 4)

    122.4\: eV

 
Answers (1)
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As we learnt in

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

 

 For Li++,  z  = 3

\Delta E=E_{0}Z^{2}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}} \right ) = 13.6\times 9\left(\frac{1}{1^{2}}-\frac{1}{3^{2}} \right ) = 13.6\times 9\times \frac{8}{9}

\Delta E=108.8 eV

Correct option is 3.


Option 1)

12.1\: eV

This is an incorrect option.

Option 2)

36.3\: eV

This is an incorrect option.

Option 3)

108.8\: eV

This is the correct option.

Option 4)

122.4\: eV

This is an incorrect option.

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