The energy of `n^(th)` orbit is given by
`E_(n) = ( -Rhc)/(n^(2))`
When electron jumps from 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)))`
The ratio of wavelength `H_(alpha)` of Lyman Series and `H_(alpha)` of Pfund Series is

Options

A. `54.7 :1`
B. `0.0183 :1`
C. `61.4 :1`
D. `0.0163 :1`

Answer

To solve the problem of finding the ratio of the wavelengths of the H-alpha line in the Lyman series and the Pfund series, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electron transitions from higher energy levels (n2) to the first energy level (n1 = 1). For the H-alpha line in the Lyman series, the transition is from n2 = 2 to n1 = 1.

### Step 2: Calculate Wavelength for Lyman Series (H-alpha)
Using the formula for wavelength:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For hydrogen (Z = 1), n1 = 1, and n2 = 2:
\[ \frac{1}{\lambda_{Lyman}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Calculating this gives:
\[ \frac{1}{\lambda_{Lyman}} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_{Lyman} = \frac{4}{3R} \]

### Step 3: Understand the Pfund Series
The Pfund series corresponds to transitions from higher energy levels to the fifth energy level (n1 = 5). For the H-alpha line in the Pfund series, the transition is from n2 = 6 to n1 = 5.

### Step 4: Calculate Wavelength for Pfund Series (H-alpha)
Using the same formula:
\[ \frac{1}{\lambda_{Pfund}} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For hydrogen (Z = 1), n1 = 5, and n2 = 6:
\[ \frac{1}{\lambda_{Pfund}} = R \left( \frac{1}{5^2} - \frac{1}{6^2} \right) \]
Calculating this gives:
\[ \frac{1}{\lambda_{Pfund}} = R \left( \frac{1}{25} - \frac{1}{36} \right) \]
Finding a common denominator (which is 900):
\[ \frac{1}{\lambda_{Pfund}} = R \left( \frac{36 - 25}{900} \right) = R \left( \frac{11}{900} \right) \]
Thus,
\[ \lambda_{Pfund} = \frac{900}{11R} \]

### Step 5: Calculate the Ratio of Wavelengths
Now, we find the ratio of the wavelengths:
\[ \frac{\lambda_{Lyman}}{\lambda_{Pfund}} = \frac{\frac{4}{3R}}{\frac{900}{11R}} = \frac{4}{3R} \times \frac{11R}{900} \]
The R cancels out:
\[ \frac{\lambda_{Lyman}}{\lambda_{Pfund}} = \frac{4 \times 11}{3 \times 900} = \frac{44}{2700} = \frac{11}{675} \]

### Final Answer
The ratio of the wavelengths of H-alpha of the Lyman series to the H-alpha of the Pfund series is:
\[ \frac{11}{675} \]