What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series ? (h = Planck constant, c = Velocity of light, R = Rydberg constant)

Options

A. `(5hcR)/(36)`
B. `(4hcR)/(3)`
C. `(3hcR)/(4)`
D. `(7hcR)/(144)`

Answer

To find the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series, we can follow these steps:

### Step 1: Identify the transition in the Lyman series
The Lyman series corresponds to electronic transitions from higher energy levels (n ≥ 2) to the first energy level (n = 1). The lowest energy transition occurs from n = 2 to n = 1.

### Step 2: Use the Rydberg formula for wavelength
The Rydberg formula for the wavelength of emitted light during these transitions is given by:

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

For the Lyman series, where \( n_1 = 1 \) and \( n_2 = 2 \):

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

### Step 3: Calculate the difference in the terms
Calculating the right-hand side:

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

### Step 4: Rearrange to find the wavelength
Thus, we have:

\[ \frac{1}{\lambda} = \frac{3R}{4} \]

This implies:

\[ \lambda = \frac{4}{3R} \]

### Step 5: Calculate the energy using the wavelength
The energy of the emitted photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Substituting for \(\lambda\):

\[ E = \frac{hc}{\frac{4}{3R}} = \frac{3hcR}{4} \]

### Step 6: Conclusion
Thus, the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series is:

\[ E = \frac{3hcR}{4} \]

### Final Answer
The correct option is \( \frac{3hcR}{4} \).

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