Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65\,?$). The de-Broglie wavelength of this electron is :

Question

The nuclides of $^{197}_{79}\mathrm{Au}$ and $^{196}_{80}\mathrm{Hg}$ are called ______ of each other

Answer 1

To find the de-Broglie wavelength of an electron in a hydrogen atom revolving in its second excited state, we first need to understand the relationship between the electron's orbit and de-Broglie wavelength.

The Bohr Model states that the electron revolves in a quantized orbit. The length of this orbit must be an integral multiple of the electron's de-Broglie wavelength:

2\pi r = n\lambda_{dB}

where r is the radius of the orbit, n is the principal quantum number, and \lambda_{dB} is the de-Broglie wavelength.

In the question, the electron is in the second excited state, which means n = 3 because the ground state is n = 1. The radius r is given as 4.65 \, \text{Å}. Our goal is to find \lambda_{dB}.

Substitute the values into the formula:

\lambda_{dB} = \frac{2\pi \times 4.65}{3} \, \text{Å}

Calculate:

\lambda_{dB} = \frac{2 \times 3.1416 \times 4.65}{3} = 9.731 \, \text{Å}

Therefore, the de-Broglie wavelength of the electron is approximately 9.7 Å, which is the correct option given.

Hence, the correct answer is 9.7 Å.

Answer 2