Question

The de-Broglie wavelength of an electron in 4th orbit is (where \( r \) = radius of the 1st orbit)

• A. \( 2\pi r \)
• B. \( 4\pi r \)
• C. \( 8\pi r \)
• D. \( 16\pi r \)

Hint:

The general formula for the de-Broglie wavelength in the \( n \)-th orbit is \( \lambda = 2\pi r n \). Since the orbit radius grows quadratically with \( n \) and the wavelength is \( \frac{2\pi r_n}{n} \), the wavelength increases linearly with \( n \) as \( \lambda_n = n \cdot \lambda_1 \). Since \( \lambda_1 = 2\pi r \), for \( n = 4 \) the wavelength is simply \( 4 \times 2\pi r = 8\pi r \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the de-Broglie wavelength \( \lambda \) of an electron orbiting in the 4th Bohr orbit of a hydrogen atom, in terms of the radius \( r \) of the 1st orbit.

Step 2: Key Formula or Approach:
1. Bohr's Quantization Condition (as a standing wave):
\[ 2\pi r_n = n \lambda \implies \lambda = \frac{2\pi r_n}{n} \]
where \( r_n \) is the radius of the \( n \)-th orbit and \( n \) is the principal quantum number.
2. Radius of the \( n \)-th Bohr orbit:
\[ r_n = r_1 \cdot n^2 = r \cdot n^2 \]

Step 3: Detailed Explanation:
1. For the 4th orbit (\( n = 4 \)):
The radius \( r_4 \) is:
\[ r_4 = r \cdot 4^2 = 16 r \]
2. Substituting \( r_4 \) and \( n = 4 \) into Bohr's quantization formula:
\[ 2\pi r_4 = 4 \lambda \]
\[ 2\pi (16 r) = 4 \lambda \]
\[ \lambda = \frac{32\pi r}{4} = 8\pi r \]

Step 4: Final Answer:
The de-Broglie wavelength of the electron in the 4th orbit is \( 8\pi r \).