Question

The doubly ionized lithium atom is hydrogen-like with the atomic number $z=3$. Find the wavelength of the radiation required to excite the electron in $L{i}^{2+}$ from the first to the third Bohr orbit is:

[Given:$\frac{1}{R_H}=912\stackrel{o}{A}]$

• A. $114\stackrel{o}{A}$
• B. $156\stackrel{o}{A}$
• C. $180\stackrel{o}{A}$
• D. $201\stackrel{o}{A}$

Answer 1

Answer: (A)

$114\stackrel{o}{A}$

Energy $=13.6Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Given, $\frac{1}{R_H}=912\stackrel{o}{A}$

$\frac{hc}{\lambda }=13.6\times Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{\lambda }=R_H\times Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$n=1$ to $n=3$

$\frac{1}{\lambda }=R_H\times 9(1-\frac{1}{9})$

$\frac{1}{\lambda }=8R_H$

$\Rightarrow \lambda =\frac{1}{8}{R}_H$

$\lambda =114\stackrel{o}{A}$

Hence, the correct option is $A$