Question

For $H$-like atoms:

$E_n=-\frac{Z^{2}Rh}{n^2};u_n=\frac{u_{1}Z}{n}$ and $r_n=\frac{r_1\times n^2}{Z};$ where $Rh$ is Rydberg.

The electrons in $L{i}^{2+}$ ions are excited from ground state by absorbing $8.4375Rh$ energy/electron. How much emission lines are expected during de-excitation of electrons to ground state?

• A. $3$
• B. $2$
• C. $6$
• D. $10$

Answer 1

Answer: (C)

$6$

For Li atom, $\Delta E=8.4375Rh,$

$E_{1_{L{i}^{2+}}}=-Rh\times 9$ and $E_{n_{L{i}^{2+}}}=\frac{-Rh\times 9}{n^2}$

$E_1-E_n=\frac{9Rh}{1}-\frac{9Rh}{n^2}=8.4375Rh$

$9-8.4375=\frac{9}{n^2}$

$n=4$

Means electron de-excite from $4^{th}$ to $1^{th}$ shell.

So, no. of spectral lines are $=\frac{n(n-1)}{2}=\frac{4\times 3}{2}=6$

Hence, option C is correct.