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:

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

Answer 2

Step 1: Recall Bohr’s quantization condition.
According to Bohr’s model, the angular momentum of an electron in the nth orbit is quantized and given by:
\[ mvr = \frac{nh}{2\pi} \] where \( m \) is the mass of the electron, \( v \) is its velocity, \( r \) is the radius of the orbit, and \( h \) is Planck’s constant.

Step 2: Relation between de-Broglie wavelength and orbit circumference.
de-Broglie proposed that the electron behaves as a standing wave in its orbit. The condition for a stable orbit is that the circumference of the orbit must be an integral multiple of the de-Broglie wavelength:
\[ 2\pi r = n\lambda \] Hence, \[ \lambda = \frac{2\pi r}{n} \]

Step 3: Substitute the radius of the nth orbit.
For a hydrogen atom, the radius of the nth orbit is given by:
\[ r_n = n^2 a_0 \] where \( a_0 \) is the Bohr radius.

Substitute \( r_n = n^2 a_0 \) into the wavelength expression:
\[ \lambda = \frac{2\pi n^2 a_0}{n} = 2\pi n a_0 \]

Step 4: For the second orbit (n = 2).
\[ \lambda = 2\pi (2)a_0 = 4\pi a_0 \] Hence, for any orbit, the de-Broglie wavelength can be written as:
\[ \lambda = \frac{4\pi a_0}{n} \]

Final Answer:
\[ \boxed{\lambda = \frac{4\pi a_0}{n}} \]