Question

An electron in a H-like atom is in an excited state. It has a total energy of 3.4 eV. Calculate the de-Broglie's wavelength.

• A. $66.5\stackrel{o}{A}$
• B. $6.66\stackrel{o}{A}$
• C. $60.6\stackrel{o}{A}$
• D. $6.06\stackrel{o}{A}$

Answer 1

The correct option is B $6.66\stackrel{o}{A}$

Total energy $=-\frac{e^2}{2r_n}=-3.4eV=\frac{E_1}{n^2}$
$\therefore n^2=\frac{-13.6}{-3.4}=4$
$n=2$
The velocity in $2^{nd}$ orbit $=\frac{v}{2}=\frac{2.18\times {10}^8}{2}\left(cmse{c}^{-1}\right)$
$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-27}\times 2}{9.1\times {10}^{-28}\times 2.18\times {10}^8}=6.6\times {10}^{-10}=6.6A^{o}A$