JEE Main 2016 · 9 Apr · Shift 1 · Physics Q22

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PhysicsSingle correct+4 / -1Question 22 of 90

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 :

Answer and worked solution

Official answer: C

The photon carries 10.2 eV. In $\ce{Li^2+}$ the level n has binding energy $\frac{13.6\times9}{n^2}$ eV, which is at most 10.2 eV when $n^2\ge12$ : the least n is 4.

Topics:Bohr model and hydrogen spectrum

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JEE Main 2016 · 9 Apr · Shift 1, Physics question 22