Energy level diagram of a certain atom is shown in the figure. The wavelength obtained in the emission transitions from level C to A is 1000 Å, and from C to B is 5000 Å. Calculate the wavelength emitted in the transition from B to A.
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The wavelength of emitted photons in atomic transitions is inversely proportional to the energy difference between the initial and final states. Smaller energy differences correspond to longer wavelengths.
Step 1: Use the energy level formula.
The energy of the photon emitted during a transition is related to the change in energy levels, and the wavelength \( \lambda \) of the emitted photon is related to the energy difference \( \Delta E \) by the formula:
\[
E = \frac{hc}{\lambda}
\]
where \( h \) is Planck’s constant and \( c \) is the speed of light.
Step 2: Energy difference for the transition C to A.
For the transition from level C to A, the wavelength is given as 1000 Å. Let \( E_C - E_A = \Delta E_{CA} \) be the energy difference between levels C and A. The energy of the photon emitted is:
\[
\Delta E_{CA} = \frac{hc}{\lambda_{CA}} = \frac{hc}{1000 \times 10^{-10}} = \frac{hc}{10^{-7} \, \text{m}}
\]
where \( \lambda_{CA} = 1000 \, \text{Å} \).
Step 3: Energy difference for the transition C to B.
Similarly, for the transition from C to B, the wavelength is 5000 Å, and the energy difference is:
\[
\Delta E_{CB} = \frac{hc}{\lambda_{CB}} = \frac{hc}{5000 \times 10^{-10}} = \frac{hc}{5 \times 10^{-7} \, \text{m}}
\]
Step 4: Energy difference for the transition B to A.
Now, the energy difference for the transition from B to A can be calculated by the relationship:
\[
\Delta E_{BA} = \Delta E_{CA} - \Delta E_{CB}
\]
Substituting the expressions for \( \Delta E_{CA} \) and \( \Delta E_{CB} \), we get:
\[
\Delta E_{BA} = \frac{hc}{10^{-7}} - \frac{hc}{5 \times 10^{-7}} = \frac{hc}{10^{-7}} \left( 1 - \frac{1}{5} \right) = \frac{hc}{10^{-7}} \times \frac{4}{5} = \frac{4hc}{5 \times 10^{-7}}
\]
Step 5: Calculate the wavelength for the transition B to A.
The wavelength for the transition from B to A is:
\[
\lambda_{BA} = \frac{hc}{\Delta E_{BA}} = \frac{hc}{\frac{4hc}{5 \times 10^{-7}}} = \frac{5 \times 10^{-7}}{4} = 1.25 \times 10^{-7} \, \text{m} = 1250 \, \text{Å}
\]
Step 6: Conclusion.
Thus, the wavelength emitted in the transition from B to A is \( 1250 \, \text{Å} \).
