Question

The wavelength corresponding to electronic transition between two orbits of hydrogen atom is $912\mathring{A}$. The wavelength (in $\mathring{A}$ ) for the same electronic transition in $L{i}^{2+}$ is

• A. 101.3
• B. 202.6
• C. 303.9
• D. 50.65

Answer 1

Answer: (A)

$\because E=\frac{hc}{\lambda }$, 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, $E\alpha Z^2\alpha \frac{l}{\lambda }$

where, $E=$ energy, $\lambda =$ wavelength

$Z=$ atomic number

Therefore, $Z_{(H)}=1$

$Z_{(Li)}=3$

$\Rightarrow {\lambda }_{(H)}=912\mathring{A}$

${\alpha }_{(Li)}=To$ find

$\therefore \frac{Z^2(H)}{Z_{(Li)}^2}=\frac{{\lambda }_{(Li)}}{{\lambda }_{(H)}}$ or ${\lambda }_{(Li)}=\frac{1\times 912}{9}$

${\lambda }_{(Li)}=101.3\mathring{A}$

Hence, wavelength $=101.3\mathring{A}$