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Solve! - Atoms And Nuclei - JEE Main

If the binding energy of the electron in a hydrogen  atom is 13.6\: eV the energy required to remove the electron from the first excited state of Li^{++} is

  • Option 1)

    30.6\: eV

  • Option 2)

    13.6\: eV

  • Option 3)

    3.4\: 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

 

 \Delta E=E_{0}z^{2} \left(\frac{1}{z^{2}}-\frac{1}{\infty^{2}} \right )=\frac{E_{0}z^{2}}{4}

\because\ \;z=3\ \; \Rightarrow\ \;\Delta E=\frac{13.6\times9}{4}=30.6\ eV

Correct option is 1.


Option 1)

30.6\: eV

This is the correct option.

Option 2)

13.6\: eV

This is an incorrect option.

Option 3)

3.4\: eV

This is an incorrect option.

Option 4)

122.4\: eV

This is an incorrect option.

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