The wave number of the first line in the Balmer series of hydrogen atom is `15200 cm^(-1)`. What is the wave number of the first line in the Balmer series of `Be^(3+)`?

Options

A. `2.43 xx 10^(5) cm^(-1)`
B. `15200 cm^(-1)`
C. `4 xx15200 cm^(-1)`
D. `2xx15200 cm^(-1)`

Answer

To find the wave number of the first line in the Balmer series of the `Be^(3+)` ion, we can use the formula for the wave number in the Balmer series:

\[ \bar{\nu} = \bar{\nu}_H \cdot \frac{Z^2}{n^2} \]

Where:
- \(\bar{\nu}\) is the wave number we want to find for `Be^(3+)`.
- \(\bar{\nu}_H\) is the wave number for hydrogen, which is given as \(15200 \, \text{cm}^{-1}\).
- \(Z\) is the atomic number of the element.
- \(n\) is the principal quantum number for the transition, which is \(2\) for the first line in the Balmer series.

### Step-by-step Solution:

1. **Identify the atomic number (Z) of `Be^(3+)`:**
- Beryllium (Be) has an atomic number of \(4\).

2. **Determine the principal quantum number (n):**
- For the first line in the Balmer series, \(n = 2\).

3. **Substitute the values into the formula:**
\[ \bar{\nu} = 15200 \, \text{cm}^{-1} \cdot \frac{Z^2}{n^2} \]
\[ \bar{\nu} = 15200 \, \text{cm}^{-1} \cdot \frac{4^2}{2^2} \]

4. **Calculate \(Z^2\) and \(n^2\):**
- \(Z^2 = 4^2 = 16\)
- \(n^2 = 2^2 = 4\)

5. **Substitute \(Z^2\) and \(n^2\) into the equation:**
\[ \bar{\nu} = 15200 \, \text{cm}^{-1} \cdot \frac{16}{4} \]

6. **Simplify the fraction:**
\[ \frac{16}{4} = 4 \]

7. **Calculate the wave number:**
\[ \bar{\nu} = 15200 \, \text{cm}^{-1} \cdot 4 = 60800 \, \text{cm}^{-1} \]

### Final Answer:
The wave number of the first line in the Balmer series of `Be^(3+)` is \(60800 \, \text{cm}^{-1}\).

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