The wavelength of the radiation emitted when an electron falls from Bohr 's orbit 4 to 2 in H atom is

Options

A. 972 nm
B. 486 nm
C. 243 nm
D. 182 nm

Answer

To find the wavelength of the radiation emitted when an electron falls from Bohr's orbit 4 to 2 in a hydrogen atom, we can follow these steps:

### Step 1: Identify the energy levels
In the Bohr model of the hydrogen atom, the energy levels are defined by the principal quantum number \( n \). Here, we have:
- \( n_1 = 2 \) (final state)
- \( n_2 = 4 \) (initial state)

### Step 2: Use the Rydberg formula
The Rydberg formula for the wavelength of emitted radiation when an electron transitions between two energy levels is given by:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).

### Step 3: Substitute the values into the formula
Substituting \( n_1 = 2 \) and \( n_2 = 4 \) into the formula:

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

Calculating the squares:

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

Finding a common denominator:

\[ \frac{1}{\lambda} = R_H \left( \frac{4 - 1}{16} \right) = R_H \left( \frac{3}{16} \right) \]

### Step 4: Calculate the wavelength
Now substituting the value of \( R_H \):

\[ \frac{1}{\lambda} = \left( 1.097 \times 10^7 \, \text{m}^{-1} \right) \left( \frac{3}{16} \right) \]

Calculating \( \frac{3}{16} \):

\[ \frac{3}{16} = 0.1875 \]

Now, substituting this back:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1875 \]

Calculating this gives:

\[ \frac{1}{\lambda} \approx 2.060625 \times 10^6 \, \text{m}^{-1} \]

Taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{2.060625 \times 10^6} \approx 4.85 \times 10^{-7} \, \text{m} = 485 \, \text{nm} \]

### Final Answer
The wavelength of the radiation emitted when an electron falls from Bohr's orbit 4 to 2 in a hydrogen atom is approximately **485 nm**.

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