An electron makes a transition from orbit `n = 4` to the orbit `n = 2` of a hydrogen atom. The wave length of the emitted radiations `(R =` Rydberg's constant) will be

Options

A. `16/(4R)`
B. `16/(5R)`
C. `16/(2R)`
D. `16/(3R)`

Answer

To find the wavelength of the emitted radiation when an electron transitions from orbit \( n = 4 \) to orbit \( n = 2 \) in a hydrogen atom, we can use the Rydberg formula for hydrogen spectral lines:

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

Where:
- \( \lambda \) is the wavelength of the emitted radiation,
- \( R \) is the Rydberg constant,
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

### Step 1: Identify the values of \( n_f \) and \( n_i \)
From the problem statement:
- The initial state \( n_i = 4 \)
- The final state \( n_f = 2 \)

### Step 2: Substitute the values into the Rydberg formula
Substituting \( n_f \) and \( n_i \) into the formula gives:

\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

### Step 3: Calculate \( \frac{1}{n_f^2} \) and \( \frac{1}{n_i^2} \)
Calculating \( \frac{1}{n_f^2} \) and \( \frac{1}{n_i^2} \):

\[ \frac{1}{n_f^2} = \frac{1}{2^2} = \frac{1}{4} \]
\[ \frac{1}{n_i^2} = \frac{1}{4^2} = \frac{1}{16} \]

### Step 4: Substitute these values back into the equation
Now substituting these values back into the equation:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

### Step 5: Find a common denominator and simplify
To simplify \( \frac{1}{4} - \frac{1}{16} \), we need a common denominator:

\[ \frac{1}{4} = \frac{4}{16} \]
So,

\[ \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]

### Step 6: Substitute back into the equation
Now substituting this back into the equation:

\[ \frac{1}{\lambda} = R \cdot \frac{3}{16} \]

### Step 7: Take the reciprocal to find \( \lambda \)
Taking the reciprocal gives:

\[ \lambda = \frac{16}{3R} \]

### Final Result
Thus, the wavelength of the emitted radiation when the electron transitions from \( n = 4 \) to \( n = 2 \) is:

\[ \lambda = \frac{16}{3R} \]