Question

Radius of the first orbit in H-atom is \(a_0\). Then, de Broglie wavelength of electron in the third orbit is.

• A. 3\(\pi\)a₀
• B. 6\(\pi\)a₀
• C. 9\(\pi\)a₀
• D. 12\(\pi\)a₀

Answer 1

The Correct Option is B

To determine the de Broglie wavelength of the electron in the third orbit of the hydrogen atom, we begin by using de Broglie's hypothesis, which states that an electron orbiting a nucleus could be treated as a wave, with its wavelength given by:

\(\lambda = \frac{h}{mv}\)

For an electron in a circular orbit, de Broglie's hypothesis leads to the condition for stable orbit, which is:

\(n\lambda = 2\pi r_n\)

where \( n \) is the principal quantum number and \( r_n \) is the radius of the \( n \)-th orbit.

From Bohr's model, the radius of the \( n \)-th orbit (\( r_n \)) is given by:

\(r_n = n^2 a_0\)

For the third orbit (\( n = 3 \)),

\(r_3 = 3^2 a_0 = 9 a_0\)

Substituting back into the condition for stable orbit, we get:

\(3\lambda = 2\pi \times 9a_0\)

Simplifying, we find the wavelength:

\(\lambda = \frac{2\pi \times 9a_0}{3} = 6\pi a_0\)

Thus, the de Broglie wavelength of the electron in the third orbit is 6\(\pi\)a₀.