Question

In terms of Planck's constant (h), permittivity of free space \((\epsilon_{0})\), mass of the electron (m) and charge of the electron (e), the de Broglie wavelength associated with the electron in the second orbit of hydrogen atom is

• A. \(\frac{h^{2}\epsilon_{0}}{2me^{2}}\)
• B. \(\frac{h^{2}\epsilon_{0}}{4me^{2}}\)
• C. \(\frac{2h^{2}\epsilon_{0}}{me^{2}}\)
• D. \(\frac{4h^{2}\epsilon_{0}}{me^{2}}\) \bigskip

Hint:

For the \(n^{th}\) orbit, \[ \lambda_n=\frac{2\pi r_n}{n} \] Using Bohr radius expressions directly saves a lot of calculation time in competitive examinations.

Answer 1

The Correct Option is D

Concept: According to Bohr's atomic model, \[ 2\pi r_n=n\lambda \] where \(\lambda\) is the de Broglie wavelength associated with the electron. Also, \[ r_n=\frac{n^2h^2\epsilon_0}{\pi me^2} \] for hydrogen atom. Using these relations we can determine the wavelength corresponding to the second orbit.

Step 1: Write the radius of the second Bohr orbit.
For \(n=2\), \[ r_2=\frac{4h^2\epsilon_0}{\pi me^2} \]

Step 2: Apply Bohr's standing wave condition.
\[ 2\pi r_n=n\lambda \] For \(n=2\), \[ 2\pi r_2=2\lambda \] \[ \lambda=\pi r_2 \] Substituting \(r_2\), \[ \lambda =\pi\left( \frac{4h^2\epsilon_0}{\pi me^2} \right) \] \[ \lambda = \frac{4h^2\epsilon_0}{me^2} \]

Step 3: Write the final answer.
Hence, \[ \boxed{\lambda= \frac{4h^2\epsilon_0}{me^2}} \]