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

The correct option is (B): \(6\pi a_0\)
 λ = \(\frac{4}{mv}\)
 λ =\(\frac{2\pi r}{n}\)
 λ = \(\frac{2\pi a_0n^2}{n}\)
 λ =\(2\pi a_0n\)
 λ = \(6\pi a_0\)

Answer 2

The de Broglie wavelength of an electron in the nth orbit of hydrogen atom can be calculated using the formula:
\(λ = \frac{h}{ p}\)
where λ is the de Broglie wavelength, h is the Planck constant, and p is the momentum of the electron.
The momentum of the electron in the nth orbit can be calculated using the Bohr model formula:
p = mv = \(\frac{nh}{2πr}\)
where m is the mass of the electron, v is its velocity, n is the principal quantum number, h is the Planck constant, and r is the radius of the nth orbit.
For the third orbit, n = 3, so the radius is:
r = 3²a₀ = 9a₀
The momentum is:
p = \(\frac{3h}{2π(9a0)}\)
The de Broglie wavelength is:
\(λ = \frac{h}{ p}\) = \(\frac{2π(9a_0)}{ 3}\)
Simplifying this expression gives:
λ = 6πa₀
Therefore, the correct answer is (B) 6\(\pi\)a₀.
Answer. B