$$ \text{An electron in the ground state of the hydrogen atom has the orbital radius of }5.3\times 10^{-11} m \text{ while that for the electron in third excited state is } 8.48\times 10^{-10}m. \text{ The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is }$$

We need to find the ratio of de Broglie wavelengths of the electron in the third excited state (n=4) to the ground state (n=1) of hydrogen.

The de Broglie wavelength is related to the orbital parameters by the Bohr quantization condition:

$$n\lambda = 2\pi r_n$$

Therefore, the de Broglie wavelength is:

$$\lambda_n = \frac{2\pi r_n}{n}$$

For the ground state (n=1): $$r_1 = 5.3 \times 10^{-11}$$ m

$$\lambda_1 = \frac{2\pi \times 5.3 \times 10^{-11}}{1} = 2\pi \times 5.3 \times 10^{-11}$$

For the third excited state (n=4): $$r_4 = 8.48 \times 10^{-10}$$ m

$$\lambda_4 = \frac{2\pi \times 8.48 \times 10^{-10}}{4} = 2\pi \times 2.12 \times 10^{-10}$$

The ratio of wavelengths:

$$\frac{\lambda_4}{\lambda_1} = \frac{2.12 \times 10^{-10}}{5.3 \times 10^{-11}} = \frac{2.12}{0.53} = 4$$

The correct answer is Option 4: 4.