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Step-by-step Solution
Step 1: Calculate the Energy of the Incident Photon
The given wavelength of the monochromatic radiation is
$ \lambda = 975 \,\text{\AA} $. We use the relation
$ E = \frac{hc}{\lambda} $ in electronvolts (eV).
Numerically, this is often converted using:
$ E(\text{eV}) = \frac{12375}{\lambda(\text{\AA})}. $
Therefore,
$ E = \frac{12375}{975} \approx 12.7\,\text{eV}. $
Step 2: Determine the Final Excited State
For a hydrogen atom initially in its ground state ($ n = 1 $),
if it absorbs a photon of about $ 12.75\,\text{eV} $,
the electron transitions to the energy level where the difference
from the ground state is about $ 12.75\,\text{eV} $.
The energy difference between $ n = 1 $ and $ n = 4 $ is
$ E_1 - E_4 = 13.6\,\text{eV} - 0.85\,\text{eV} = 12.75\,\text{eV}. $
Hence, the electron reaches the fourth energy level ($ n = 4 $).
Step 3: Find the Number of Spectral Lines Emitted
The electron can drop from $ n = 4 $ back to lower energy levels
through various pathways, emitting photons of corresponding
frequencies. The total number of distinct spectral lines
for an electron dropping from level
$ n $ to the ground state is given by the formula:
$ \frac{n(n - 1)}{2}. $
Here, $ n = 4 $. Thus,
$ \frac{4 \times (4 - 1)}{2} = \frac{4 \times 3}{2} = 6. $
Step 4: Conclusion
The hydrogen atom emits 6 distinct spectral lines
when the electron transitions from the fourth energy level
down to the ground state in all possible ways.
