Question

Electronic transitions in $L{i}^{2+}$ ions take place from $n_2$ to $n_1$ shell such that:$2n_2+3n_1=18$ and $2n_2-3n_1=6$. The total number of photons emitted when electrons transit to $n_1$ shell is x (electrons do not fall lower than $n_1$ shell) and number of photons emitted in the Balmer range is y. Find value of $x+y$.

Answer 1

Given,
$2n_2+3n_1=18$
$2n_2-3n_1=6$
Solving these two equations, we get $n_1=2$, $n_2=6$.
So total number of photons when electrons transit to $n_1$ shell will be $\left(x\right)=\frac{(n_2-n_1+1)(n_2-n_1)}{2}=\frac{(6-2+1)(6-2)}{2}=10$
Photons in Balmer range (visible range) will be $6\to 2,5\to 2,4\to 2,3\to 2$
So, number of photons in Balmer range will be $\left(y\right)=4$
$x+y=14$