The frequency of 1st line Balmer series in `H_(2)` atom is `v_(0)`. The frequency of line emitted by single ionised He atom is

Options

A. `2v_(0)Hz`
B. `4v_(0)Hz`
C. `(v_(0)//2)Hz`
D. `(v_(0)//4)Hz`

Answer

To solve the problem, we need to find the frequency of the line emitted by a singly ionized helium atom (He\(^+\)) in relation to the frequency of the first line of the Balmer series in a hydrogen atom (H). The frequency of the first line of the Balmer series in hydrogen is given as \( v_0 \).

### Step-by-Step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to the transitions of electrons in hydrogen from higher energy levels to the second energy level (n=2). The first line of this series is the transition from n=3 to n=2.

2. **Identify the Atomic Numbers**:
- For hydrogen (H), the atomic number \( Z_1 = 1 \).
- For singly ionized helium (He\(^+\)), the atomic number \( Z_2 = 2 \).

3. **Frequency Relation**:
The frequency of emitted light is proportional to the square of the atomic number. Therefore, we can express the relationship between the frequencies of hydrogen and helium as:
\[ \frac{v_1}{v_2} = \frac{Z_1^2}{Z_2^2} \]
where \( v_1 \) is the frequency for hydrogen and \( v_2 \) is the frequency for helium.

4. **Substituting Values**:
Substitute the atomic numbers into the equation:
\[ \frac{v_0}{v_2} = \frac{1^2}{2^2} = \frac{1}{4} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ v_2 = 4v_0 \]

6. **Conclusion**:
Thus, the frequency of the line emitted by the singly ionized helium atom is:
\[ v_2 = 4v_0 \]

### Final Answer:
The frequency of the line emitted by the singly ionized helium atom is \( 4v_0 \).

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