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