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 $Li ^{2+}$ is

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

Answer 1

Answer: (A)

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

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 $=101.3\, \mathring{A}$