Detection of Hydrogen
Hydrogen is prevalent in the universe. Life in the universe is ultimately based on hydrogen.
The electronic energy of a hydrogen atom is given by `DeltaE(n_(1) rarr n)=-C(1//n_(1)^(2)-1//n_(1)^(2))`, relative to zero energy at infinite separation between the electron and the proton (n is the principle quantum number, and C is a constant). For detection of the `DeltaE=(3 rarr 2)` transition `(656.3 nm` in the Balmer series), the electron in the ground state of the hydrogen atom needs to be excited first to the absorption line in the starlight corresponding to the `DeltaE=(1 rarr 2)` transition.

Answer

To solve the problem of detecting hydrogen through its electronic transitions, we will follow these steps:

### Step 1: Understand the Energy Transition Formula
The electronic energy of a hydrogen atom is given by the formula:
\[ \Delta E(n_1 \rightarrow n) = -C \left( \frac{1}{n_1^2} - \frac{1}{n^2} \right) \]
where \( n_1 \) is the initial principal quantum number, \( n \) is the final principal quantum number, and \( C \) is a constant.

### Step 2: Calculate the Energy for the Transition from n=3 to n=2
For the transition from \( n=3 \) to \( n=2 \):
- Set \( n_1 = 3 \) and \( n = 2 \):
\[ \Delta E(3 \rightarrow 2) = -C \left( \frac{1}{3^2} - \frac{1}{2^2} \right) = -C \left( \frac{1}{9} - \frac{1}{4} \right) \]
Calculating the fractions:
\[ \frac{1}{9} - \frac{1}{4} = \frac{4 - 9}{36} = -\frac{5}{36} \]
Thus,
\[ \Delta E(3 \rightarrow 2) = -C \left( -\frac{5}{36} \right) = \frac{5C}{36} \]

### Step 3: Calculate the Wavelength for the Transition
The wavelength \( \lambda \) corresponding to this energy transition can be calculated using the formula:
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting \( \Delta E \):
\[ \lambda = \frac{hc}{\frac{5C}{36}} = \frac{36hc}{5C} = \frac{36h}{5} \cdot \frac{1}{c} \]
Given that \( \lambda \) for this transition is approximately \( 656.3 \, \text{nm} \).

### Step 4: Calculate the Energy for the Transition from n=1 to n=2
Now, we calculate the energy for the transition from \( n=1 \) to \( n=2 \):
- Set \( n_1 = 1 \) and \( n = 2 \):
\[ \Delta E(1 \rightarrow 2) = -C \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = -C \left( 1 - \frac{1}{4} \right) = -C \left( \frac{3}{4} \right) \]
Thus,
\[ \Delta E(1 \rightarrow 2) = \frac{3C}{4} \]

### Step 5: Calculate the Wavelength for the Transition from n=1 to n=2
Using the same wavelength formula:
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting \( \Delta E \):
\[ \lambda = \frac{hc}{\frac{3C}{4}} = \frac{4hc}{3C} \]
To find the wavelength, we can relate it to the previous transition:
\[ \lambda_{1 \rightarrow 2} = \frac{656.3 \, \text{nm} \cdot \frac{5C}{36}}{\frac{3C}{4}} = 656.3 \cdot \frac{5 \cdot 4}{36 \cdot 3} \]
Calculating this gives:
\[ \lambda_{1 \rightarrow 2} = 121.5 \, \text{nm} \]

### Final Answer
The wavelength for the transition from \( n=1 \) to \( n=2 \) is \( 121.5 \, \text{nm} \).

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