Question

A hydrogen atom makes a transition from $n=2$ to $n=1$ and emits a photon. This photon strikes a doubly ionized lithium atom $(z=3)$ in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is

• A. $3$
• B. $2$
• C. $5$
• D. $4$

Answer 1

The correct option is D $4$

For this to occur, energy of transition must be more than ionization energy of lithium, hence,

$1-\frac{1}{2^2}>3^2\frac{1}{n^2}$

$n>\sqrt{12}$

Hence, $n_{min}=4$.