The wave number of the first emission line in the Balmer series of H-Spectrum is : (R = Rydberg constant)

The wave number for a spectral line in the hydrogen spectrum is given by the Rydberg formula:

$$\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

where $$\bar{\nu}$$ is the wave number, $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the higher energy level.

For the Balmer series, the electron transitions end at the $$n = 2$$ level. Therefore, $$n_1 = 2$$. The first emission line corresponds to the smallest energy transition, which is from the next higher level, $$n_2 = 3$$, down to $$n_1 = 2$$.

Substituting $$n_1 = 2$$ and $$n_2 = 3$$ into the formula:

$$\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$

Calculate the squares: $$2^2 = 4$$ and $$3^2 = 9$$, so:

$$\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{9} \right)$$

To subtract these fractions, find a common denominator. The least common multiple of 4 and 9 is 36. Rewrite each fraction:

$$\frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36}$$

Now subtract:

$$\frac{9}{36} - \frac{4}{36} = \frac{5}{36}$$

Therefore:

$$\bar{\nu} = R \times \frac{5}{36} = \frac{5}{36}R$$

Comparing with the options, $$\frac{5}{36}R$$ corresponds to option A.

Hence, the correct answer is Option A.