Question

The de-Broglie's wavelength of an electron in the \( 4^{\text{th}} \) orbit is ______ \( \pi a_0 \). (\( a_0 = \text{Bohr's radius} \))

Answer 1

Correct Answer: 8

The condition for an electron's de-Broglie wavelength in an orbit is stated as:

\[ 2 \pi r_n = n \lambda_d \]

Here, \( r_n \) denotes the radius of the \( n \)-th orbit, \( n \) is the principal quantum number, and \( \lambda_d \) represents the de-Broglie wavelength.

The radius of the \( n \)-th orbit within the Bohr model is defined by:

\[ r_n = 2 \pi a_0 \frac{n^2}{Z} \]

Substituting this expression yields:

\[ 2 \pi a_0 \frac{n^2}{Z} = n \lambda_d \]

For an electron occupying the 4th orbit of a hydrogen atom (\( Z = 1 \)):

\[ 2 \pi a_0 \frac{4^2}{1} = 4 \lambda_d \]

Upon simplification:

\[ \lambda_d = 8 \pi a_0 \]

Therefore, the de-Broglie wavelength of the electron in the 4th orbit is determined to be \( 8 \pi a_0 \).