The atomic spectrum of hydrogen is found to contain a series of lines of wavelengths 656.46, 486.27, 434.17 and 410.29 nm. The wavelength (in nm) of the next line in the series will be....

Explanation

To find the next wavelength in the series of hydrogen's atomic spectrum, we can follow these steps:

### Step 1: Identify the Series
The given wavelengths are part of the Balmer series, which corresponds to transitions of electrons from higher energy levels (n ≥ 3) to the second energy level (n = 2). The wavelengths provided are:
- 656.46 nm (n=3 to n=2)
- 486.27 nm (n=4 to n=2)
- 434.17 nm (n=5 to n=2)
- 410.29 nm (n=6 to n=2)

### Step 2: Determine the Next Transition
The next transition in the Balmer series would be from n = 7 to n = 2.

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelengths of spectral lines in hydrogen is given by:

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

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 \) is the lower energy level (2 for Balmer series)
- \( n_2 \) is the higher energy level (7 for the next line)

### Step 4: Calculate the Wavelength for n=7 to n=2
Substituting \( n_1 = 2 \) and \( n_2 = 7 \):

\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{7^2} \right) \]

Calculating the values:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{49} \right) \]

Calculating \( \frac{1}{4} - \frac{1}{49} \):

\[ \frac{1}{4} = 0.25 \]
\[ \frac{1}{49} \approx 0.0204 \]
\[ \frac{1}{4} - \frac{1}{49} = 0.25 - 0.0204 = 0.2296 \]

Now substituting back into the formula:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.2296 \approx 2.52 \times 10^6 \]

### Step 5: Calculate the Wavelength
Now, taking the reciprocal to find \( \lambda \):

\[ \lambda \approx \frac{1}{2.52 \times 10^6} \approx 3.97 \times 10^{-7} \text{ m} = 397 \text{ nm} \]

### Conclusion
The wavelength of the next line in the series will be approximately **397 nm**.

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