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 hydrogen. It states that electrons occupy specific energy levels (orbits) and can transition between these levels by absorbing or emitting energy in the form of photons.

2. **Wavelength Formula**:
The wavelength of the emitted photon when an electron transitions from a higher energy level (n_initial) to a lower energy level (n_final) can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \]
where \( R \) is the Rydberg constant.

3. **Identifying the Series**:
Different series correspond to different values of \( n_{final} \):
- Lyman series: \( n_{final} = 1 \)
- Balmer series: \( n_{final} = 2 \)
- Paschen series: \( n_{final} = 3 \)

4. **Finding the Shortest Wavelength**:
The shortest wavelength corresponds to the maximum energy transition. For a given series, the shortest wavelength occurs when \( n_{initial} \) is at its maximum value. In theory, this means \( n_{initial} \) approaches infinity.

5. **Setting \( n_{initial} \) to Infinity**:
When \( n_{initial} \) approaches infinity, the term \( \frac{1}{n_{initial}^2} \) approaches zero. Thus, the formula simplifies:
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{n_{final}^2} - 0 \right) = \frac{R}{n_{final}^2} \]
Therefore, the shortest wavelength is given by:
\[ \lambda_{min} = \frac{n_{final}^2}{R} \]

6. **Conclusion**:
The shortest wavelength in a series is determined by the lower energy level \( n_{final} \) into which the electron falls. The higher the \( n_{final} \), the longer the wavelength, and vice versa. Thus, the shortest wavelength corresponds to \( n_{final} = 1 \) for the Lyman series.

### Final Answer:
The shortest wavelength in a given series of wavelengths emitted by the atom is determined by the quantum number \( n_{final} \), which identifies the lower energy level into which the electron falls.