Using Bohr's model , calculate the wavelength of the radiation emitted when an electron in a hydrogen atom makes a transition from the fourth energy level to the second energy level.

Explanation

To solve the problem of calculating the wavelength of radiation emitted when an electron in a hydrogen atom transitions from the fourth energy level (n1 = 4) to the second energy level (n2 = 2) using Bohr's model, we can follow these steps:

### Step 1: Understand the Formula
The formula we will use is derived from the Rydberg formula for hydrogen:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
Where:
- \(\lambda\) is the wavelength of the emitted radiation.
- \(R_H\) is the Rydberg constant (\(1.1 \times 10^7 \, \text{m}^{-1}\) for hydrogen).
- \(Z\) is the atomic number (for hydrogen, \(Z = 1\)).
- \(n_1\) is the lower energy level (4 in this case).
- \(n_2\) is the higher energy level (2 in this case).

### Step 2: Substitute the Values
Substituting the known values into the formula:
\[ \frac{1}{\lambda} = R_H \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \]

### Step 3: Calculate the Terms Inside the Parentheses
Calculating the terms:
\[ \frac{1}{4} = \frac{4}{16} \]
So,
\[ \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]

### Step 4: Substitute Back into the Formula
Now substituting back into the equation:
\[ \frac{1}{\lambda} = R_H \cdot \frac{3}{16} \]
Substituting the value of \(R_H\):
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \cdot \frac{3}{16} \]

### Step 5: Calculate \(\frac{1}{\lambda}\)
Calculating the right-hand side:
\[ \frac{1}{\lambda} = \frac{3.3 \times 10^7}{16} \]
Calculating this gives:
\[ \frac{1}{\lambda} = 2.0625 \times 10^6 \, \text{m}^{-1} \]

### Step 6: Calculate \(\lambda\)
To find \(\lambda\), take the reciprocal:
\[ \lambda = \frac{1}{2.0625 \times 10^6} \approx 4.85 \times 10^{-7} \, \text{m} \]
Converting this to nanometers (1 m = \(10^9\) nm):
\[ \lambda \approx 485 \, \text{nm} \]

### Final Answer
The wavelength of the radiation emitted when an electron in a hydrogen atom transitions from the fourth energy level to the second energy level is approximately **485 nm**.

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