Question

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength ${\lambda }_1:{\lambda }_2$emitted in the two cases is

• A. $7/5$
• B. $27/20$
• C. $27/5$
• D. $20/7$

Answer 1

Answer: (D)

Here, for wavelength ${\lambda }_1$
$n_1=4$ and $n_2=3$
and for ${\lambda }_2,n_1=3$ and $n_2=2$
We have $\frac{hc}{\lambda }=-13.6\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$
so, for ${\lambda }_1$
$\Rightarrow \frac{hc}{{\lambda }_1}=-13.6\left[\frac{1}{(4{)}^2}-\frac{1}{(3{)}^2}\right]$
$\frac{hc}{{\lambda }_1}=-13.6\left[\frac{7}{144}\right]\dots$ (i)
simiarly, for ${\lambda }_2$
$\Rightarrow \frac{hc}{{\lambda }_1}=-13.6\left[\frac{1}{(3{)}^2}-\frac{1}{(2{)}^2}\right]$
$\frac{hc}{{\lambda }_1}=-13.6\left[\frac{5}{36}\right]\dots$ (ii)
Hence, from Eqs. (i) and (ii), we get
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{-13.6\left[\frac{7}{144}\right]}{-13.6\left[\frac{5}{36}\right]}=\frac{20}{7}$