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-by-Step Solution:

1. **Identify the Initial and Final Energy Levels:**
- The electron falls from n2 = 4 to n1 = 2.

2. **Use the Rydberg Formula:**
- The formula for the wavelength (λ) of the emitted radiation 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 \( 1.0967 \times 10^7 \, \text{m}^{-1} \).

3. **Substitute the Values:**
- Here, \( n_1 = 2 \) and \( n_2 = 4 \).
- Substitute these values into the formula:
\[ \frac{1}{\lambda} = 1.0967 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

4. **Calculate the Terms:**
- Calculate \( \frac{1}{2^2} = \frac{1}{4} = 0.25 \)
- Calculate \( \frac{1}{4^2} = \frac{1}{16} = 0.0625 \)
- Now, substitute these into the equation:
\[ \frac{1}{\lambda} = 1.0967 \times 10^7 \left( 0.25 - 0.0625 \right) \]
- Simplifying the terms gives:
\[ 0.25 - 0.0625 = 0.1875 \]

5. **Calculate \( \frac{1}{\lambda} \):**
- Now calculate:
\[ \frac{1}{\lambda} = 1.0967 \times 10^7 \times 0.1875 \]
- This results in:
\[ \frac{1}{\lambda} \approx 2.058125 \times 10^6 \, \text{m}^{-1} \]

6. **Find the Wavelength \( \lambda \):**
- To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{2.058125 \times 10^6} \approx 486 \times 10^{-9} \, \text{m} \]

7. **Final Result:**
- Therefore, the wavelength of the radiation emitted is approximately \( 486 \, \text{nm} \).