The first member of the Balmer series of hydrogen atom has a wavelength of 6561 $$\mathring{A}$$. The wavelength of the second member of the Balmer series (in nm) is

We have to find the wavelength of the second member of the Balmer series of the hydrogen atom, given that the first member has a wavelength of 6561 Å.

First, we convert the given wavelength from angstrom to nanometre because the final answer is required in nm. We know that

$$1\;\text{\AA}=0.1\;\text{nm}.$$

So,

$$\lambda_1 = 6561\;\text{\AA}=6561\times0.1\;\text{nm}=656.1\;\text{nm}.$$

For hydrogen spectral lines we use the Rydberg formula. We state it first:

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

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level and $$n_2$$ is the higher energy level with $$n_2>n_1$$.

In the Balmer series $$n_1=2$$ and $$n_2=3,4,5,\ldots$$ 

The first member (H-α line) corresponds to $$n_2=3$$. Therefore, using the given wavelength $$\lambda_1$$, we write

$$\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) =R\left(\frac{1}{4}-\frac{1}{9}\right).$$

Simplifying the bracket:

$$\frac{1}{4}=\frac{9}{36},\qquad \frac{1}{9}=\frac{4}{36},\qquad \frac{1}{4}-\frac{1}{9}=\frac{9}{36}-\frac{4}{36}=\frac{5}{36}.$$

So,

$$\frac{1}{\lambda_1}=R\left(\frac{5}{36}\right).$$

Re-arranging to obtain $$R$$, we get

$$R=\frac{1}{\lambda_1}\cdot\frac{36}{5}.$$

Now we move to the second member (H-β line) of the Balmer series, for which $$n_2=4$$. Again using the Rydberg formula:

$$\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) =R\left(\frac{1}{4}-\frac{1}{16}\right).$$

Simplifying inside the bracket:

$$\frac{1}{4}=\frac{4}{16},\qquad \frac{1}{16}=\frac{1}{16},\qquad \frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}.$$

Hence,

$$\frac{1}{\lambda_2}=R\left(\frac{3}{16}\right).$$

Substituting the expression of $$R$$ obtained from the first line:

$$\frac{1}{\lambda_2}=\left(\frac{1}{\lambda_1}\cdot\frac{36}{5}\right)\left(\frac{3}{16}\right).$$

Multiplying the numerical factors:

$$\frac{36}{5}\times\frac{3}{16}=\frac{36\times3}{5\times16} =\frac{108}{80}=\frac{27}{20}.$$

Thus,

$$\frac{1}{\lambda_2}=\frac{1}{\lambda_1}\cdot\frac{27}{20}.$$

Taking the reciprocal of both sides to isolate $$\lambda_2$$:

$$\lambda_2=\lambda_1\cdot\frac{20}{27}.$$

Now we substitute $$\lambda_1=656.1\;\text{nm}$$:

$$\lambda_2=656.1\;\text{nm}\times\frac{20}{27}.$$

Carrying out the multiplication-division step by step:

$$\frac{20}{27}\approx0.740740\dots$$

$$\lambda_2=656.1\times0.740740\dots\;\text{nm}\approx486.0\;\text{nm}.$$

So, the wavelength of the second member of the Balmer series is $$486\;\text{nm}$$.

Hence, the correct answer is Option C.