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 \( \boxed{4} \).

Answer 2

In the Bohr model, the radius of the \(n^{th}\) orbit is given by:

\[ r_n = n^2 r_1 \]

Given \(r_4 = 8.48\times10^{-10}\) and \(r_1 = 5.3\times10^{-11}\):

\[ \frac{r_4}{r_1} = \frac{8.48\times10^{-10}}{5.3\times10^{-11}} \approx 16 \]

Thus, \(r_4 = 16r_1 \Rightarrow n^2 = 16 \Rightarrow n = 4.\)

The de Broglie wavelength of the electron in the \(n^{th}\) orbit is related by Bohr’s quantization condition:

\[ 2\pi r_n = n\lambda_n \Rightarrow \lambda_n = \frac{2\pi r_n}{n} \]

Therefore, the ratio of wavelengths in ground and fourth states is:

\[ \frac{\lambda_1}{\lambda_4} = \frac{r_1/n_1}{r_4/n_4} = \frac{r_1}{r_4}\cdot\frac{n_4}{n_1} \] \[ \frac{\lambda_1}{\lambda_4} = \frac{1}{16}\times4 = \frac{1}{4} \]

Hence, the ratio of de Broglie wavelengths of the electron in the ground state to that in the excited state is:

\[ \boxed{\frac{\lambda_1}{\lambda_4} = \frac{1}{4}} \]

Or equivalently, \(\lambda_4 : \lambda_1 = 4 : 1\).

Answer

Option \(\lambda_4 : \lambda_1 = 4 : 1\).