calculate the longest wavelength in Balmer series and the series limit . (Given `R=1.097xx10^(7)m^(-1)` )

Options

A. 6000 A
B. 6563 A
C. 7200 A
D. 7383 A

Answer

To calculate the longest wavelength in the Balmer series and the series limit, we can use the Rydberg formula for hydrogen. The formula is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, m^{-1}\)),
- \(n_1\) is the lower energy level,
- \(n_2\) is the higher energy level.

### Step 1: Determine the values of \(n_1\) and \(n_2\) for the longest wavelength in the Balmer series.
The Balmer series corresponds to transitions where \(n_1 = 2\). The longest wavelength occurs when the transition is from the next higher level, which is \(n_2 = 3\).

### Step 2: Substitute the values into the Rydberg formula.
Using \(n_1 = 2\) and \(n_2 = 3\):

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

Calculating the squares:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 3: Find a common denominator and simplify.
The common denominator of 4 and 9 is 36:

\[ \frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 4: Rearrange to find \(\lambda\).
Thus, we have:

\[ \lambda = \frac{36}{5R} \]

### Step 5: Substitute the value of \(R\).
Substituting \(R = 1.097 \times 10^7 \, m^{-1}\):

\[ \lambda = \frac{36}{5 \times 1.097 \times 10^7} \]

Calculating the denominator:

\[ 5 \times 1.097 \times 10^7 = 5.485 \times 10^7 \]

Now calculating \(\lambda\):

\[ \lambda = \frac{36}{5.485 \times 10^7} \approx 6.56 \times 10^{-7} \, m \]

### Step 6: Convert to Angstroms.
To convert meters to Angstroms (1 Angstrom = \(1 \times 10^{-10} \, m\)):

\[ \lambda \approx 6.56 \times 10^{-7} \, m \times \frac{1 \times 10^{10} \, \text{Angstrom}}{1 \, m} \approx 6563 \, \text{Angstrom} \]

### Step 7: Calculate the series limit.
For the series limit, we take \(n_2\) to be infinite:

\[ \frac{1}{\lambda_{limit}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]

Thus:

\[ \lambda_{limit} = \frac{4}{R} \]

Substituting \(R\):

\[ \lambda_{limit} = \frac{4}{1.097 \times 10^7} \approx 3.64 \times 10^{-7} \, m \]

Converting to Angstroms:

\[ \lambda_{limit} \approx 3.64 \times 10^{-7} \, m \times \frac{1 \times 10^{10} \, \text{Angstrom}}{1 \, m} \approx 3640 \, \text{Angstrom} \]

### Final Answers:
- Longest wavelength in the Balmer series: **6563 Angstrom**
- Series limit: **3640 Angstrom**