The energy of `n^(th)` orbit is given by
`E_(n) = ( -Rhc)/(n^(2))`
When electron jumpsfrom one orbit to another orbit then wavelength associated with the radiation is given by
`(1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/ (n_(2)^(2)))`
When electron of 1.0 gm atom of Hydrogen undergoes transition giving the spectral line of lowest energy in visible region of its atomic spectra, the wavelength of radiation is

Options

A. `6564Å`
B. `3640Å`
C. `5000Å`
D. `4312Å`

Answer

To solve the problem, we need to determine the wavelength of radiation emitted when an electron in a hydrogen atom transitions between energy levels, specifically focusing on the lowest energy spectral line in the visible region.

### Step-by-Step Solution:

1. **Identify the Transition**:
The lowest energy transition in the visible region for hydrogen occurs in the Balmer series. The transition that gives the lowest energy corresponds to an electron moving from the third energy level (n2 = 3) to the second energy level (n1 = 2).

2. **Use the Wavelength Formula**:
The formula for the wavelength associated with the transition is given by:
\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (\( R = 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 = 2 \) and \( n_2 = 3 \)

3. **Substitute Values into the Formula**:
Plugging in the values, we have:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \]

4. **Calculate the Difference**:
To find \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Thus,
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

5. **Substitute Back into the Wavelength Formula**:
Now substituting back:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} \]

6. **Calculate \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} = \frac{1.097 \times 10^7 \times 5}{36} \]
Performing the multiplication:
\[ \frac{1}{\lambda} = \frac{5.485 \times 10^7}{36} \approx 1.524 \times 10^6 \, \text{m}^{-1} \]

7. **Calculate \( \lambda \)**:
Now take the reciprocal to find \( \lambda \):
\[ \lambda = \frac{1}{1.524 \times 10^6} \approx 6.56 \times 10^{-7} \, \text{m} \]
Converting to angstroms (1 m = \( 10^{10} \) angstroms):
\[ \lambda \approx 6.56 \times 10^{-7} \times 10^{10} \approx 6560 \, \text{angstroms} \]

8. **Final Answer**:
Rounding to the nearest whole number gives us:
\[ \lambda \approx 6564 \, \text{angstroms} \]

Thus, the wavelength of radiation emitted when the electron transitions from n=3 to n=2 in a hydrogen atom is **6564 angstroms**.