The shortest wavelength which can be obtained in hydrogen spectrum `(R=10 ^(7) m^(-1))`

Options

A. 1000 Å
B. 8000 Å
C. 1300 Å
D. 2100 Å

Explanation

To find the shortest wavelength that can be obtained in the hydrogen spectrum, we can use the Rydberg formula:

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

Where:
- \( \lambda \) is the wavelength,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( R \) is the Rydberg constant, and
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron's energy levels.

### Step 1: Identify the values for \( n_1 \) and \( n_2 \)
To find the shortest wavelength, we want to maximize the term \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \). This occurs when:
- \( n_1 = 1 \) (the ground state),
- \( n_2 \) approaches infinity (the ionization limit).

### Step 2: Substitute the values into the Rydberg formula
Substituting \( Z = 1 \), \( n_1 = 1 \), and \( n_2 = \infty \) into the Rydberg formula gives:

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

This simplifies to:

\[ \frac{1}{\lambda} = R \left( 1 - 0 \right) = R \]

### Step 3: Calculate \( \lambda \)
Now, we know that \( R = 10^7 \, \text{m}^{-1} \). Therefore:

\[ \frac{1}{\lambda} = 10^7 \, \text{m}^{-1} \]

Taking the reciprocal gives:

\[ \lambda = \frac{1}{10^7} \, \text{m} = 10^{-7} \, \text{m} \]

### Step 4: Convert to Angstroms
To convert meters to Angstroms (where \( 1 \, \text{Angstrom} = 10^{-10} \, \text{m} \)):

\[ \lambda = 10^{-7} \, \text{m} = 10^{-7} \times \frac{1 \, \text{Angstrom}}{10^{-10} \, \text{m}} = 1000 \, \text{Angstroms} \]

### Final Answer
The shortest wavelength that can be obtained in the hydrogen spectrum is:

\[ \lambda = 1000 \, \text{Angstroms} \]

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