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

 

Answers (1)

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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