The first emission line of Balmer series in `He^(+)` - spectrum has the wave no .in `(Cm^(_1))` (R-Rydberg constant)

Options

A. `(3R)/(4)`
B. `(20R)/(36)`
C. `(5R)/(36)`
D. `(R)/(6)`

Answer

To find the wave number of the first emission line of the Balmer series in the He\(^+\) spectrum, we can follow these steps:

### Step 1: Identify the relevant quantum numbers
For the first emission line of the Balmer series, we need to identify the initial and final energy levels. The Balmer series corresponds to transitions where the final energy level \( n_1 \) is 2, and the initial energy level \( n_2 \) is 3.

- \( n_1 = 2 \)
- \( n_2 = 3 \)

### Step 2: Use the formula for wave number
The wave number \( \bar{\nu} \) (in cm\(^{-1}\)) for the transition can be calculated using the Rydberg formula:

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

Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for He\(^+\), \( Z = 2 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers.

### Step 3: Substitute the values into the formula
Substituting the values into the formula:

\[ \bar{\nu} = R \cdot (2^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating \( Z^2 \):

\[ Z^2 = 2^2 = 4 \]

Now calculate \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \):

\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} \]

Finding a common denominator (36):

\[ \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 4: Combine the results
Now substitute back into the wave number equation:

\[ \bar{\nu} = R \cdot 4 \cdot \frac{5}{36} \]

This simplifies to:

\[ \bar{\nu} = \frac{20}{36} R \]

### Step 5: Simplify the fraction
We can simplify \( \frac{20}{36} \):

\[ \bar{\nu} = \frac{5}{9} R \]

### Final Answer
Thus, the wave number of the first emission line of the Balmer series in the He\(^+\) spectrum is:

\[ \bar{\nu} = \frac{5}{9} R \text{ (in cm}^{-1}\text{)} \]