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Step-by-Step Detailed Solution
Step 1: Understand the Energy Transition
In a hydrogen-like atom, when an electron transitions between energy levels (shells), it emits (or absorbs) radiation of a wavelength that depends on the difference in the energy levels. Here, the shells are labeled as K (n=1), L (n=2), M (n=3), and N (n=4).
Step 2: Write Down the Formula for Wavelength
For a hydrogen-like atom, the reciprocal of the wavelength (often called the wavenumber) for a transition from an initial energy level $n_i$ to a final energy level $n_f$ is given by:
$ \frac{1}{\lambda} = K \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) $,
where $K$ is a constant (proportional to the Rydberg constant for the atom in question).
Step 3: Apply the Formula for the M → L Transition
The M-shell corresponds to $n = 3$ and the L-shell corresponds to $n = 2$. Hence, for the transition from M to L:
$ \frac{1}{\lambda} = K \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
= K \left( \frac{1}{4} - \frac{1}{9} \right)
= K \left( \frac{9 - 4}{36} \right)
= \frac{5K}{36}. $
This defines $ \lambda $, the wavelength of the photon emitted during the M → L transition.
Step 4: Apply the Formula for the N → L Transition
The N-shell corresponds to $n = 4$ and again the L-shell is $n = 2$. Thus, for the transition from N to L:
$ \frac{1}{\lambda'} = K \left( \frac{1}{2^2} - \frac{1}{4^2} \right)
= K \left( \frac{1}{4} - \frac{1}{16} \right)
= K \left( \frac{4 - 1}{16} \right)
= \frac{3K}{16}. $
This defines $ \lambda' $, the wavelength of the photon emitted during the N → L transition.
Step 5: Relate the Two Wavelengths
From above, we have:
$ \frac{1}{\lambda} = \frac{5K}{36} \quad \text{and} \quad \frac{1}{\lambda'} = \frac{3K}{16}. $
Rewriting $ \lambda $ and $ \lambda' $ in terms of these expressions:
$ \lambda = \frac{36}{5K}, \quad \lambda' = \frac{16}{3K}. $
Now, to find $ \frac{\lambda'}{\lambda} $:
$ \frac{\lambda'}{\lambda}
= \frac{\frac{16}{3K}}{\frac{36}{5K}}
= \frac{16}{3K} \times \frac{5K}{36}
= \frac{16 \times 5}{3 \times 36}
= \frac{80}{108} = \frac{20}{27}.
$
Hence, $ \lambda' = \frac{20}{27} \lambda. $
Step 6: Conclude the Correct Answer
Therefore, the wavelength of the photon emitted when the electron jumps from the N-shell to the L-shell is
$ \frac{20}{27} \lambda. $
This matches the correct option: $ \frac{20}{27} \lambda. $
