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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 an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for this process is?

5

3

2

4

Solution:

For this to occur, the energy of the transition must be more than the ionization energy of lithium. Hence,

13.6(1/1² - 1/2²) > 13.6 * 3²/n²

13.6(1 - 1/4) > 13.6 * 9/n²

3/4 > 9/n²

n² > 12

Hence, nmin = 4.