# 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

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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