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 $\frac{h c}{\lambda}=-13.6\left[\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right]$
so, for $\lambda_{1}$
$\Rightarrow \frac{h c}{\lambda_{1}}=-13.6\left[\frac{1}{(4)^{2}}-\frac{1}{(3)^{2}}\right]$
$\frac{h c}{\lambda_{1}}=-13.6\left[\frac{7}{144}\right] \ldots$ (i)
simiarly, for $\lambda_{2}$
$\Rightarrow \frac{h c}{\lambda_{1}}=-13.6\left[\frac{1}{(3)^{2}}-\frac{1}{(2)^{2}}\right]$
$\frac{h c}{\lambda_{1}}=-13.6\left[\frac{5}{36}\right] \ldots$ (ii)
Hence, from Eqs. (i) and (ii), we get
$\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{-13.6\left[\frac{7}{144}\right]}{-13.6\left[\frac{5}{36}\right]}=\frac{20}{7}$