Question

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

• A. 4
• B. 9
• C. 3
• D. 16

Hint:

For electrons in orbit, the de Broglie wavelength is inversely proportional to the orbital radius.

Answer 1

The Correct Option is A

The de Broglie wavelength is related to the radius \( r \) by the following formula: \[ \lambda = \frac{h}{mv} \] For circular orbits, we also have: \[ mv r = \frac{nh}{2\pi} \] Hence, the wavelength \( \lambda \) is inversely proportional to the radius \( r \): \[ \lambda \propto \frac{1}{r} \] The ratio of the wavelengths in the ground state and the third excited state is: \[ \frac{\lambda_1}{\lambda_4} = \frac{r_1n_4}{r_4n_1} = \frac{5.3 \times 10^{-11} \times 4}{1 \times 8.48 \times 10^{-10}} = \frac{4}{1} \] Thus, the ratio is \( \boxed{4} \).