# The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy? Option 1) $I$ Option 2) $II$ Option 3) $III$ Option 4) $IV$

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

$R= R hydberg\: constant$

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

Highest difference of energy is between n = 1 and n = 3.

According to $\Delta E=E_{0}z^{2} \left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}} \right )$. Correct option is III. Since I is absorption of photon.

Correct option is 3.

Option 1)

$I$

This is an incorrect option.

Option 2)

$II$

This is an incorrect option.

Option 3)

$III$

This is the correct option.

Option 4)

$IV$

This is an incorrect option.

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