Q1 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Chemistry
The wavelength corresponding to electronic transition between two orbits of hydrogen atom is $912 Å$. The wavelength (in $Å$ ) for the same electronic transition in $\mathrm{Li}^{2+}$ is
- Correct answerA101.3
- B202.6
- C303.9
- D50.65
Solution
$\because E=\frac{h c}{\lambda}$, i.e. $E \alpha \frac{1}{\lambda}$ and $E \alpha \frac{Z^2}{n^2}$
For same value of (n) $E \alpha Z^2$
Thus, $E \alpha Z^2 \alpha \frac{1}{\lambda}$
where, $E=$ energy, $\lambda=$ wavelength
$Z=$ atomic number
Therefore, $Z_{(\mathrm{H})}=1$
$$
\begin{aligned}
Z_{(\mathrm{Li})} & =3 \quad \Rightarrow \quad \lambda_{(\mathrm{H})}=912 Å \\
\alpha_{(\mathrm{Li})} & =\text { To find } \\
\therefore \quad \frac{Z_{(\mathrm{H})}^2}{Z_{(\mathrm{Li})}} & =\frac{\lambda_{(\mathrm{Li})}}{\lambda_{(\mathrm{H})}} \text { or } \lambda_{(\mathrm{Li})}=\frac{1 \times 912}{9} \\
\lambda_{(\mathrm{Li})} & =101.3 Å
\end{aligned}
$$
Hence, wavelength $=101.3 Å$
Hence, the correct answer is option: 'A'.