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 the hydrogen atom?

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

Hint:

For electrons in orbits, the de-Broglie wavelength is related to the orbit radius and quantum number. Higher \( n \) results in a larger wavelength.

Answer 1

The Correct Option is C

The de-Broglie wavelength of an electron in orbit is given by: \[ \lambda = \frac{h}{mv} \] For the electron in the second orbit, the wavelength is inversely proportional to the quantum number \( n \), and can be expressed as: \[ \lambda = \frac{4\pi a_0}{n} \] Thus, the correct expression for the wavelength is \( \frac{4\pi a_0}{n} \).