The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively
Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to `(n(n - 1))/(2)`If teh electron comes back from the energy level having energy `E_(2)` to the energy level having energy `E_(1)` then the difference may be expresent in terms of energy of photon as `E_(2) - E_(1) = Delta E, lambda = hc//Delta E` Since h and c are constants `Delta E` coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula `bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2)))`Where R is a Rydherg constant
The difference in the wavelength of the second line is Lyman series and last line of breaker series is a hydrogen sample is

Options

A. `(119)/(8R)`
B. `(1271)/(8R)`
C. `(219)/(8R)`
D. None of these

Explanation

To solve the problem, we need to find the difference in the wavelengths of the second line of the Lyman series and the last line of the Balmer series for a hydrogen atom. We will use the Rydberg formula for hydrogen spectral lines.

### Step-by-Step Solution:

1. **Identify the Energy Levels for the Series:**
- For the **Lyman series**, the transitions occur from higher energy levels down to the first level (n=1). The second line corresponds to a transition from n=3 to n=1.
- For the **Balmer series**, the transitions occur from higher energy levels down to the second level (n=2). The last line corresponds to a transition from n=4 to n=2.

2. **Write the Rydberg Formula:**
The Rydberg formula for the wavelength (λ) of emitted light is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (1 for hydrogen), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Calculate Wavelength for the Lyman Series:**
- For the second line of the Lyman series:
- \( n_1 = 1 \) (ground state)
- \( n_2 = 3 \)
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]
\[ \lambda_L = \frac{9}{8R} \]

4. **Calculate Wavelength for the Balmer Series:**
- For the last line of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 4 \)
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
\[ \lambda_B = \frac{4}{R} \]

5. **Calculate the Difference in Wavelengths:**
Now, we need to find the difference in wavelengths:
\[ \Delta \lambda = \lambda_B - \lambda_L = \frac{4}{R} - \frac{9}{8R} \]
To combine these fractions, we need a common denominator:
\[ \Delta \lambda = \frac{32}{8R} - \frac{9}{8R} = \frac{32 - 9}{8R} = \frac{23}{8R} \]

### Final Answer:
The difference in the wavelength of the second line of the Lyman series and the last line of the Balmer series in a hydrogen sample is:
\[ \Delta \lambda = \frac{23}{8R} \]