Question

Calculate the ratio of the energies of the $n=3$ to $n=1$ transitions of the hydrogen atom to the $L{i}^{2+}$ ion.

• A. 1 : 3
• B. 1 : 9
• C. 3 : 1
• D. 9 : 1

Answer 1

The correct option is B 1 : 9

The energy transition is given by $E\left(eV\right)=-13.6Z^2(\frac{1}{n_{f^2}}-\frac{1}{n_{i^2}})$
Since, both 'n' values for the transitions in question are the same the ratio of the energies is determined only by the ratio of the $Z^2$ values or $\frac{(1{)}^2}{(3{)}^2}=\frac{1}{9}$. Thus, the energy of the $n=3$ to $n=1$ transition in $L{i}^{2+}$ ion is 9 times greater than that for ${}^{\prime }{H}^{\prime }$ atom.