In balmer series of emission spectrum of hydrogen, first four lines with different wavelength `H_alpha`, `H_beta`, `H_gamma` and `H_delta` are obtained. Which line has maximum frequency out of these ?

Options

A. 1. `H_alpha`
B. 2. `H_beta`
C. 3. `H_gamma`
D. 4. `H_delta`

Explanation

To determine which line in the Balmer series of the hydrogen emission spectrum has the maximum frequency, we can follow these steps:

### Step 1: Understand the Wavelengths of the Lines
The first four lines of the Balmer series are:
- \( H_\alpha \): 656.3 nm
- \( H_\beta \): 486.1 nm
- \( H_\gamma \): 434.1 nm
- \( H_\delta \): 410.2 nm

### Step 2: Recall the Relationship Between Frequency and Wavelength
The frequency (\( f \)) of light is related to its wavelength (\( \lambda \)) by the equation:
\[ f = \frac{c}{\lambda} \]
where \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s). This means that frequency is inversely proportional to wavelength.

### Step 3: Analyze the Wavelengths
To find which line has the maximum frequency, we need to identify which wavelength is the shortest, since a shorter wavelength corresponds to a higher frequency.

### Step 4: Compare the Wavelengths
From the given wavelengths:
- \( H_\alpha \): 656.3 nm
- \( H_\beta \): 486.1 nm
- \( H_\gamma \): 434.1 nm
- \( H_\delta \): 410.2 nm

The shortest wavelength is \( H_\delta \) at 410.2 nm.

### Step 5: Conclusion
Since \( H_\delta \) has the shortest wavelength, it will have the maximum frequency. Therefore, the line with the maximum frequency in the Balmer series is:
\[ H_\delta \]

### Summary
The line with the maximum frequency in the Balmer series of the hydrogen emission spectrum is \( H_\delta \).

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