Question

The de-Broglie wavelength $\left({\lambda }_B\right)$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $\left({\lambda }_G\right)$ by :

• A. ${\lambda }_B={\lambda }_G/3$
• B. ${\lambda }_B={\lambda }_G/2$
• C. ${\lambda }_B=2{\lambda }_G$
• D. ${\lambda }_B=3{\lambda }_G$

Answer 1

The correct option is D ${\lambda }_B=3{\lambda }_G$

de- Broglie wavelength $\lambda =\frac{h}{mv}$

By conservation of angular momentum, $mvr=\frac{nh}{2\pi }$

$\frac{h}{mv}=\frac{2\pi r}{n}$

$\lambda =\frac{2\pi r}{n}$

$r=a_0\frac{n^2}{Z}$

$\lambda =\frac{2\pi a_{0}n}{Z}$

As the atom is hydrogen $Z=1$

${\lambda }_B=\frac{2\pi a_{0}3}{Z}$

${\lambda }_G=\frac{2\pi a_0}{Z}$

${\lambda }_B=3{\lambda }_G$