Q2 / 180 · KCET 2010 · Chemistry
The wave number of the spectral line in the emission spectrum of hydrogen will be equal to $\frac{8}{9}$ times the Rydberg's constant if the electron jumps from
- Correct answerA$\mathrm{n}=3$ to $\mathrm{n}=1$
- B$\mathrm{n}=10$ to $\mathrm{n}=1$
- C$\mathrm{n}=9$ to $\mathrm{n}=1$
- D$\mathrm{n}=2$ to $\mathrm{n}=1$
Solution
Wave number of spectral line in emission spectrum of hydrogen,
$$
\bar{v}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \quad \text{...(i)}
$$
Given, $\quad \bar{v}=\frac{8}{9} R_{H}$
On putting the value of $\bar{v}$ in Eq. (i), we get
$$
\begin{aligned}
\frac{8}{9} \mathrm{R}_{\mathrm{H}} &=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right) \\
\frac{8}{9} &=\frac{1}{(1)^{2}}-\frac{1}{\mathrm{n}_{2}^{2}} \\
\frac{8}{9}-1 &=-\frac{1}{\mathrm{n}_{2}^{2}} \\
\frac{1}{3} &=\frac{1}{\mathrm{n}_{2}} \\
\therefore \quad \mathrm{n}_{2} &=3
\end{aligned}
$$
Hence, electron jumps from $\mathrm{n}_{2}=3$ to $\mathrm{n}_{1}=1$
Hence, the correct answer is option: 'A'.