What is the colour corresponding to the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from n = 4 to n = 2?

Options

A. Blue
B. Red
C. Yellow
D. Green

Explanation

To find the color corresponding to the wavelength of light emitted when the electron in a hydrogen atom transitions from n = 4 to n = 2, we can follow these steps:

### Step 1: Identify the Transition
The electron transitions from the initial energy level \( n_i = 4 \) to the final energy level \( n_f = 2 \).

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength \( \lambda \) of light emitted during an electron transition is given by:

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

where \( R_H \) is the Rydberg constant, approximately \( 109677 \, \text{cm}^{-1} \).

### Step 3: Substitute the Values
Substituting \( n_f = 2 \) and \( n_i = 4 \) into the formula:

\[ \frac{1}{\lambda} = 109677 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the fractions:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{4^2} = \frac{1}{16} = 0.0625 \]

Now, substituting these values back into the equation:

\[ \frac{1}{\lambda} = 109677 \left( 0.25 - 0.0625 \right) = 109677 \left( 0.1875 \right) \]

### Step 4: Calculate \( \frac{1}{\lambda} \)
Calculating \( 109677 \times 0.1875 \):

\[ \frac{1}{\lambda} \approx 20564.4 \, \text{cm}^{-1} \]

### Step 5: Find \( \lambda \)
Now, we can find \( \lambda \):

\[ \lambda = \frac{1}{20564.4} \, \text{cm} \]

Calculating this gives:

\[ \lambda \approx 4.86 \times 10^{-5} \, \text{cm} \]

### Step 6: Convert to Meters and Nanometers
Convert \( \lambda \) from centimeters to meters:

\[ \lambda = 4.86 \times 10^{-5} \, \text{cm} = 4.86 \times 10^{-7} \, \text{m} \]

Now, convert to nanometers (1 nm = \( 10^{-9} \) m):

\[ \lambda = 4.86 \times 10^{-7} \, \text{m} = 486 \, \text{nm} \]

### Step 7: Determine the Color
The wavelength of \( 486 \, \text{nm} \) corresponds to blue light.

### Final Answer
The color corresponding to the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from \( n = 4 \) to \( n = 2 \) is **blue light**.

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