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 the de-Broglie wavelength of an electron in an orbit is given by:

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

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

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

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

Substituting this into the equation:

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

For an electron in the 4th orbit and for a hydrogen atom (\( Z = 1 \)):

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

Simplifying:

\[ \lambda_d = 8 \pi a_0 \]

Hence, the de-Broglie's wavelength of the electron in the 4th orbit is \( 8 \pi a_0 \).