What transition in `He^(o+)` ion shall have the same wave number as the first line in Balmer series of H atom ?

Options

A. `7 rarr 5`
B. `6 rarr 4`
C. `5 rarr 3`
D. `4 rarr 2`

Explanation

To solve the problem of determining what transition in the `He^(o+)` ion has the same wave number as the first line in the Balmer series of the hydrogen atom, we can follow these steps:

### Step 1: Determine the wave number of the first line in the Balmer series of hydrogen.
The wave number (ṽ) for the Balmer series can be calculated using the Rydberg formula:

\[ \tilde{\nu} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For the first line in the Balmer series:
- \( n_1 = 2 \) (the lower energy level)
- \( n_2 = 3 \) (the higher energy level)

Substituting these values into the formula gives:

\[ \tilde{\nu} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]

Thus,

\[ \tilde{\nu} = R_H \left( \frac{9 - 4}{36} \right) = R_H \cdot \frac{5}{36} \]

### Step 2: Set up the equation for the `He^(o+)` ion.
For the `He^(o+)` ion (which has \( Z = 2 \)), the wave number can also be expressed as:

\[ \tilde{\nu} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting \( Z = 2 \):

\[ \tilde{\nu} = R_H \cdot 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 3: Equate the two wave numbers.
We want the wave number from the `He^(o+)` ion to equal that of the hydrogen atom:

\[ R_H \cdot \frac{5}{36} = R_H \cdot 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Dividing both sides by \( R_H \):

\[ \frac{5}{36} = 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Solve for \( n_1 \) and \( n_2 \).
Rearranging gives:

\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{5}{144} \]

To find suitable values for \( n_1 \) and \( n_2 \), we can try pairs of integers. A reasonable guess is \( n_1 = 4 \) and \( n_2 = 6 \):

Calculating:

\[ \frac{1}{4^2} - \frac{1}{6^2} = \frac{1}{16} - \frac{1}{36} \]

Finding a common denominator (144):

\[ \frac{1}{16} = \frac{9}{144}, \quad \frac{1}{36} = \frac{4}{144} \]

Thus,

\[ \frac{9}{144} - \frac{4}{144} = \frac{5}{144} \]

This confirms that the transition from \( n = 6 \) to \( n = 4 \) in the `He^(o+)` ion has the same wave number as the first line in the Balmer series of hydrogen.

### Final Answer:
The transition in `He^(o+)` that has the same wave number as the first line in the Balmer series of the hydrogen atom is from \( n = 6 \) to \( n = 4 \).

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