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

Explanation

To solve the problem, we need to determine the frequency of the first line of the Balmer series for a singly ionized helium atom (He⁺) based on the frequency of the first line of the Balmer series in a hydrogen atom (H₂) given as \( v_0 \).

### Step-by-Step Solution:

1. **Understanding the Balmer Series**:
The Balmer series corresponds to transitions of electrons in hydrogen (and hydrogen-like atoms) from higher energy levels to the second energy level (n=2). The first line of the Balmer series corresponds to the transition from n=3 to n=2.

2. **Frequency Relation in Hydrogen**:
The frequency of the emitted photon during this transition can be expressed using the Rydberg formula:
\[ v = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
where \( R_H \) is the Rydberg constant for hydrogen, \( n_1 = 2 \), and \( n_2 = 3 \).

3. **Ionized Helium Atom (He⁺)**:
The singly ionized helium atom (He⁺) is a hydrogen-like atom with \( Z = 2 \) (where Z is the atomic number). The frequency of the emitted photon for He⁺ can be similarly expressed:
\[ v_{He^+} = R_{He} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
where \( R_{He} = Z^2 R_H = 2^2 R_H = 4 R_H \).

4. **Calculating the Frequency for He⁺**:
For the transition from n=3 to n=2 in He⁺, we have:
\[ v_{He^+} = 4 R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
This can be simplified to:
\[ v_{He^+} = 4 R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
To combine the fractions:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ v_{He^+} = 4 R_H \cdot \frac{5}{36} = \frac{20 R_H}{36} = \frac{5 R_H}{9} \]

5. **Relating to \( v_0 \)**:
Since we know that the frequency of the first line of the Balmer series in hydrogen is given as \( v_0 \):
\[ v_0 = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \cdot \frac{5}{36} \]
Therefore, we can express \( R_H \) in terms of \( v_0 \):
\[ R_H = \frac{36 v_0}{5} \]

6. **Substituting \( R_H \) back**:
Now substituting this back into the frequency for He⁺:
\[ v_{He^+} = \frac{5}{9} \cdot \frac{36 v_0}{5} = \frac{36 v_0}{9} = 4 v_0 \]

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