A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon.  This photon strikes a doubly ionized lithium atom  (z = 3) in excited state and completely removes the orbiting electron.  The least quantum number for the excited state of the ion for the process is :

  • Option 1)

     2

  • Option 2)

    3

  • Option 3)

     4

  • Option 4)

    5

 

Answers (1)
P perimeter

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

 

 Energy emitted by Hydrogen atom =1.36 \left(1-\frac{1}{4} \right )=10.2\ eV

So 10.2 eV photon removes an electron from nth orbital of Lithium.

Hence, Energy of nth orbital of Lithium should be - 10.2 eV.

\because\ \;E_{n}.\frac{z^{2}}{n^{2}}=-13.6\times\frac{9}{n^{2}}=-10.2

n^{2}\geq\frac{-13.6\times 9}{-10.2}=12

n\geq\sqrt{12}

Hence, Value of n = 4

Correct option is 3


Option 1)

 2

This is an incorrect option.

Option 2)

3

This is an incorrect option.

Option 3)

 4

This is the correct option.

Option 4)

5

This is an incorrect option.

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