Question

The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $(\lambda_G)$ by :

• A. $\lambda_B = 2 \lambda_G$
• B. $\lambda_B = 3 \lambda_G$
• C. $\lambda_B = \lambda_{G/2}$
• D. $\lambda_B = 2 \lambda_{G/3}$

Answer 1

Answer: (B)

We know that, $\lambda=\frac{h}{m v}$
From third Bohr's postulate, we have
$m v r=n \frac{h}{2 \pi}$
$\frac{h}{m v}=\frac{2 \pi r}{n}$
$\Rightarrow \lambda=\frac{2 \pi r}{n}$
Since $r=a_{0} \frac{n^{2}}{Z}$, where $a_{0}$ is radius of Bohr's orbit having value $(0.53 \mathring{A})1^{2}=0.53 \mathring{A}$,
therefore, $\lambda=\frac{2 \pi a_{0} n^{2}}{n \cdot Z}=\frac{2 \pi a_{0}}{Z} \cdot n$
For Hydrogen, $Z=1$. Therefore,
$\lambda_{ G }=\frac{2 \pi a_{0} \times 1}{1}=2 \pi a_{0}$
and $\lambda_{ B }=\frac{2 \pi a_{0} \times 3}{1}=6 \pi a_{0}$
Then $\lambda_{ B }=3 \lambda_{G}$