Question

The $E_1$ of $H$ is $13.6eV$. It is exposed to electromagnetic waves of $1028\mathring{A}$ and gives out induced radiation. Find the wavelengths of these induced radiation.

• A. ${\lambda }_3=1216\mathring{A}$
• B. ${\lambda }_2=6568\mathring{A}$
• C. ${\lambda }_2=1000\mathring{A}$
• D. ${\lambda }_2=2432\mathring{A}$

Answer 1

Answer: (A), (B)

${\lambda }_2=6568\mathring{A}$
${\lambda }_3=1216\mathring{A}$

$E_1$ for $H$-atom $=-13.6eV$
$\because E=\frac{12375}{\lambda }$; when $\lambda$ is in $\mathring{A}$
$\therefore$ Energy given to $H$-atom $=\frac{12375}{1028}eV=12.07eV$
$\therefore$ Energy of $H$-atom after excitation $=-13.6+12.07=-1.53eV$
$\because E_n=\frac{E_1}{n^2}$
$\therefore n^2=\frac{-13.6}{-1.53}=9;$ $\therefore n=3$
That is electrons are excited to III level. The de-excitation leads to following $\lambda$ expressed by:
$\frac{12375}{{\lambda }_1}=E_3-E_1=-1.53-(-13.6)eV$
$\therefore {\lambda }_1=1028\mathring{A}$
$\frac{12375}{{\lambda }_2}=E_3-E_2=-1.53-\frac{-13.6}{4}eV$
$\therefore {\lambda }_2=6568\mathring{A}$
$\frac{12375}{{\lambda }_3}=E_2-E_1=-\frac{13.6}{4}-(-13.6)eV$
$\therefore {\lambda }_3=1216\mathring{A}$.
Hence options A & B are correct.