In each of the following atoms or ions, electronic transition form `n=4 to n=1` take place. Frequency of the radiation emitted out will be minimum for

Options

A. Hydrogen atom
B. Deuterium atom
C. `He^(+)` ion
D. `Li^(2+)` ion

Answer

To solve the question regarding the frequency of radiation emitted during electronic transitions from \( n=4 \) to \( n=1 \) in different atoms or ions, we can follow these steps:

### Step 1: Understand the Formula
The frequency of radiation emitted during an electronic transition can be expressed using the formula:
\[ f = R \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where:
- \( f \) is the frequency of the emitted radiation,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number of the atom or ion,
- \( n_f \) is the final energy level (1 in this case),
- \( n_i \) is the initial energy level (4 in this case).

### Step 2: Identify the Values
In this case, we are transitioning from \( n_i = 4 \) to \( n_f = 1 \). The formula simplifies to:
\[ f = R \cdot Z^2 \left( 1 - \frac{1}{16} \right) = R \cdot Z^2 \cdot \frac{15}{16} \]

### Step 3: Determine the Minimum Frequency
From the formula, we can see that the frequency \( f \) is directly proportional to \( Z^2 \). Therefore, to minimize the frequency, we need to minimize \( Z \).

### Step 4: Identify the Atom with Minimum Z
The atomic number \( Z \) is the smallest for hydrogen, which has \( Z = 1 \). Other atoms or ions will have higher atomic numbers, leading to higher frequencies.

### Conclusion
Thus, the frequency of the radiation emitted will be minimum for the hydrogen atom (where \( Z = 1 \)).

### Final Answer
The frequency of the radiation emitted will be minimum for hydrogen (option A).

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