Question

If \( a_0 \) is denoted as the Bohr radius of the hydrogen atom, then what is the de-Broglie wavelength \( \lambda \) of the electron present in the second orbit of hydrogen atom? (\( n \): any integer)

• A. \( \frac{4n a_0}{\pi a_0} \)
• B. \( \frac{8 \pi a_0}{n} \)
• C. \( \frac{4 \pi a_0}{n} \)
• D. \( \frac{2 a_0}{n \pi} \)

Hint:

For Bohr orbits, the de-Broglie wavelength is inversely related to the principal quantum number \( n \) and directly proportional to the Bohr radius.

Answer 1

The Correct Option is C

The de-Broglie wavelength \( \lambda \) for an electron in an orbit of radius \( r_n \) is given by: \[ \lambda = \frac{h}{mv} \] For the electron in the second orbit of hydrogen, the radius \( r_n \) is proportional to \( n^2 \), and using Bohr's model, the wavelength for the second orbit \( n = 2 \) is given by: \[ \lambda = \frac{4 \pi a_0}{n} \] Thus, the correct answer is \( \frac{4 \pi a_0}{n} \).