According to the Bohr model , what determines the shortest wavelength in a given series of wavelength emitted by the atom ?

Options

A. The quantum number `n_i` that identifies the higher energy level from which the electron falls into a lower energy level.
B. The quantum number `n_f` that identifies the lower energy level into which the electron falls from a higher energy level.
C. The ratio `n_f//n_i` where `n_f` is the quantum number that identifies the lower energy level into which the electron falls and `n_i` is the quantum number that identifies the higher level from which the electron falls .
D. The sum `n_f +n_f` of two quantum numbers, where `n_f` identifies the lower energy `n_i` identifies the higher level from which the electron falls .

Answer

To determine the shortest wavelength in a given series of wavelengths emitted by an atom according to the Bohr model, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Bohr Model**: The Bohr model describes the behavior of electrons in an atom, particularly how they transition between different energy levels. When an electron transitions from a higher energy level (initial state, \( n_i \)) to a lower energy level (final state, \( n_f \)), it emits a photon whose wavelength can be calculated.

2. **Wavelength Formula**: The relationship between the wavelength (\( \lambda \)) of the emitted photon and the energy levels is given by the formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( n_i \) is the initial energy level, and \( n_f \) is the final energy level.

3. **Maximizing \( \frac{1}{\lambda} \)**: To find the shortest wavelength, we need to maximize the term \( \frac{1}{\lambda} \). This occurs when the term \( \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \) is maximized.

4. **Setting \( n_f \) to Infinity**: The maximum value of \( \frac{1}{n_f^2} \) occurs when \( n_f \) approaches infinity. In this case, \( \frac{1}{n_f^2} \) approaches 0, making the term \( \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \) equal to \( -\frac{1}{n_i^2} \). Thus, the equation simplifies to:
\[ \frac{1}{\lambda} = R \left( 0 - \frac{1}{n_i^2} \right) = -\frac{R}{n_i^2} \]
This indicates that as \( n_f \) approaches infinity, the wavelength \( \lambda \) becomes the shortest possible.

5. **Conclusion**: Therefore, the shortest wavelength in a given series is determined by the final energy level \( n_f \) being set to infinity. The initial energy level \( n_i \) only determines which series of spectral lines we are observing (e.g., Lyman, Balmer series, etc.).

### Final Answer:
The shortest wavelength in a given series of wavelengths emitted by the atom is determined by setting the final energy level \( n_f \) to infinity.