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, $\lambda =\frac{h}{mv}$
From third Bohr's postulate, we have
$mvr=n\frac{h}{2\pi }$
$\frac{h}{mv}=\frac{2\pi r}{n}$
$\Rightarrow \lambda =\frac{2\pi r}{n}$
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 ${\lambda }_B=3{\lambda }_G$