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

To determine the de-Broglie wavelength \(\lambda\) of the electron in the second orbit of the hydrogen atom, we start by analyzing the quantization conditions for electrons in Bohr's model and the de-Broglie wavelength relation.

The de-Broglie wavelength is given by:

\(\lambda = \frac{h}{mv}\)

where:

• \(h\) is Planck's constant.
• \(m\) is the mass of the electron.
• \(v\) is the velocity of the electron.

For an electron in the nth orbit of a hydrogen atom, the quantized angular momentum condition is:

\(mvr = \frac{nh}{2\pi}\)

Solving for \(v\), we get:

\(v = \frac{nh}{2\pi mr}\)

Substituting in the expression for the radius of the \(n\)th orbit:

\(r = n^2a_0\)

where \(a_0\) is the Bohr radius, and substituting this into the expression for \(v\):

\(v = \frac{nh}{2\pi mn^2a_0}\)

Plug this velocity \(v\) back into the de-Broglie wavelength equation:

\(\lambda = \frac{h}{m\left(\frac{nh}{2\pi mn^2a_0}\right)}\)

which simplifies to:

\(\lambda = \frac{2\pi n^2a_0}{nh}\)

The terms simplify further to:

\(\lambda = \frac{2\pi n a_0}{h}\)

In the second orbit, where \(n = 2\), we substitute \(n = 2\):

\(\lambda = \frac{4\pi a_0}{h}\)

This matches with the option:

\(\frac{4\pi a_0}{n}\)