The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $$\frac{\lambda_1}{\lambda_2}$$, of the photons emitted in this process is:

For a hydrogen atom the stationary‐state energy levels are given by the Bohr formula $$E_n=-\dfrac{13.6\ \text{eV}}{n^{2}},$$ where $$n=1,2,3,\ldots$$ is the principal quantum number.

The electron is stated to start from the third excited state. Counting from the ground state $$n=1$$, we have:

Ground state : $$n=1$$
First excited state : $$n=2$$
Second excited state : $$n=3$$
Third excited state : $$n=4$$

Thus the sequence of transitions is $$n=4 \to n=3$$ followed by $$n=3 \to n=2$$.

The energy released as a photon when the electron falls from a higher level $$n_2$$ to a lower level $$n_1$$ is obtained from the difference of the two level energies:

$$\Delta E = E_{n_1}-E_{n_2}= -\dfrac{13.6}{n_1^{2}} - \Bigl(-\dfrac{13.6}{n_2^{2}}\Bigr)=13.6\Bigl(\dfrac{1}{n_1^{2}}-\dfrac{1}{n_2^{2}}\Bigr)\ \text{eV}.$$

For the first photon of wavelength $$\lambda_1$$ corresponding to $$n=4 \to n=3$$ we have

$$\Delta E_1 = 13.6\Bigl(\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\Bigr).$$

We simplify term by term:

$$\dfrac{1}{3^{2}}=\dfrac{1}{9},\quad \dfrac{1}{4^{2}}=\dfrac{1}{16}.$$

So

$$\Delta E_1 = 13.6\Bigl(\dfrac{1}{9}-\dfrac{1}{16}\Bigr)=13.6\Bigl(\dfrac{16-9}{144}\Bigr)=13.6\times\dfrac{7}{144}\ \text{eV}.$$

For the second photon of wavelength $$\lambda_2$$ corresponding to $$n=3 \to n=2$$ we write

$$\Delta E_2 = 13.6\Bigl(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\Bigr).$$

Calculating the fractions:

$$\dfrac{1}{2^{2}}=\dfrac{1}{4},\quad \dfrac{1}{3^{2}}=\dfrac{1}{9}.$$

Hence

$$\Delta E_2 = 13.6\Bigl(\dfrac{1}{4}-\dfrac{1}{9}\Bigr)=13.6\Bigl(\dfrac{9-4}{36}\Bigr)=13.6\times\dfrac{5}{36}\ \text{eV}.$$

The energy of a photon and its wavelength are connected by the Planck relation $$E=\dfrac{hc}{\lambda}.$$ Therefore for two photons, $$\lambda\propto\dfrac{1}{E}$$, and the ratio of wavelengths is the inverse ratio of the corresponding energy differences:

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{\Delta E_2}{\Delta E_1}.$$

Substituting the expressions we have just obtained:

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{13.6\times\dfrac{5}{36}}{13.6\times\dfrac{7}{144}}.$$

The common factor $$13.6$$ cancels out, giving

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{\dfrac{5}{36}}{\dfrac{7}{144}}.$$

Dividing one fraction by another is equivalent to multiplying by its reciprocal, so

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{5}{36}\times\dfrac{144}{7}.$$

We simplify step by step. First, reduce $$144/36$$:

$$\dfrac{144}{36}=4.$$

Hence

$$\dfrac{\lambda_1}{\lambda_2}=5\times\dfrac{4}{7}=\dfrac{20}{7}.$$

Thus the required ratio of wavelengths is $$\dfrac{20}{7}.$$

Hence, the correct answer is Option B.