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

The de Broglie wavelength of an electron in a circular orbit of a hydrogen atom is given by:

$$\lambda = \frac{2\pi r}{n}$$

where:

• $r$ = orbital radius,
• $n$ = principal quantum number of the state.

Step 1: De Broglie wavelength in the ground state ($n = 1$):

For the ground state ($n = 1$), the orbital radius is:

$$r_1 = 5.3 \times 10^{-11} \, \text{m}$$

The de Broglie wavelength is:

${\lambda }_1=\frac{2\pi r_1}{1}=2\pi \cdot (5.3\times {10}^{-11})$

Step 2: De Broglie wavelength in the third excited state ($n = 4$):

For the third excited state ($n = 4$), the orbital radius is:

$$r_4 = 8.48 \times 10^{-10} \, \text{m}$$

The de Broglie wavelength is:

${\lambda }_4=\frac{2\pi r_4}{4}=\frac{2\pi \cdot (8.48\times {10}^{-10})}{4}$

Step 3: Ratio of de Broglie wavelengths:

The ratio of de Broglie wavelengths is:

$\frac{{\lambda }_1}{{\lambda }_4}=\frac{1}{4}=0.25$