Which of the following transitions give the highest frequency for electron emission?

Options

A. `n_(1)=1` to `n_(2)=2`
B. `n_(1)=2` to `n_(2)=1`
C. `n_(1)=2` to `n_(2)=5`
D. `n_(1)=5` to `n_(2)=2`

Explanation

To solve the problem of determining which transition gives the highest frequency for electron emission, we can follow these steps:

### Step 1: Understand the relationship between frequency and energy transitions
The frequency of emitted radiation during an electron transition can be derived from the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( \lambda \) is the wavelength, \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

### Step 2: Relate frequency to wavelength
The frequency \( \nu \) can be related to wavelength by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Thus, we can express frequency in terms of the Rydberg formula:
\[ \nu = R \cdot c \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

### Step 3: Identify the conditions for electron emission
For electron emission, the transition must be from a higher energy level to a lower energy level, meaning \( n_f < n_i \).

### Step 4: Evaluate the given transitions
We need to evaluate the transitions provided in the options to find which one yields the highest frequency. We will calculate \( \frac{1}{n_f^2} - \frac{1}{n_i^2} \) for each transition.

1. **Option 1:** \( n_f = 2, n_i = 1 \)
\[ \frac{1}{2^2} - \frac{1}{1^2} = \frac{1}{4} - 1 = -\frac{3}{4} \quad (\text{Not possible, negative}) \]

2. **Option 2:** \( n_f = 1, n_i = 2 \)
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \quad (\text{Positive, possible}) \]

3. **Option 3:** \( n_f = 2, n_i = 5 \)
\[ \frac{1}{2^2} - \frac{1}{5^2} = \frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100} \quad (\text{Positive, possible}) \]

4. **Option 4:** \( n_f = 4, n_i = 2 \)
\[ \frac{1}{4^2} - \frac{1}{2^2} = \frac{1}{16} - \frac{1}{4} = \frac{1 - 4}{16} = -\frac{3}{16} \quad (\text{Not possible, negative}) \]

### Step 5: Compare the positive values
Now we compare the positive values obtained from the possible transitions:
- Option 2: \( \frac{3}{4} \)
- Option 3: \( \frac{21}{100} \)

To compare, convert \( \frac{3}{4} \) to a fraction with a denominator of 100:
\[ \frac{3}{4} = \frac{75}{100} \]

### Conclusion
Since \( \frac{75}{100} > \frac{21}{100} \), the transition that gives the highest frequency for electron emission is **Option 2**.