Question

Hydrogen atoms absorb radiation of wavelength ${\lambda }_0$ and consequently emit radiations of $6$ different wavelengths of which two wavelengths are shorter than ${\lambda }_0$. Then,

• A. The final excited state of the atoms is $n=4$
• B. The initial state of the atoms may be $n=2$
• C. The initial state of the atoms may be $n=3$
• D. There are three transitions belonging to Lyman series

Answer 1

Answer: (A), (B), (D)

The final excited state of the atoms is $n=4$
The initial state of the atoms may be $n=2$
There are three transitions belonging to Lyman series

Let initial state be $n_1$
$\frac{1}{{\lambda }_o}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
ginen 6 wavelengths
${}^n{C}_2=6$
$n^2-n-12=0$
$(n-4)(n+3)=0$
$n=4or-3$
Final state = $n_2=4$
Now two wavelenghts are shorter than ${\lambda }_o$
$\frac{1}{\lambda }=R(1-\frac{1}{n_1^2})$
$R(1-\frac{1}{n_1^2})>R(\frac{1}{n_1^2}-\frac{1}{4^2})$
$n_1>2.82$
$n_1=3$
Lymen series transitions
$4\to 1,3\to 1,2\to 1$