QJEEM16Ph7d2b: JEE Main 2016 Physics - Class 12 JEEM16Ph7d2b Solution: A hydrogen atom makes a transition from n = 2 to n = 1 and…

Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,

E₁ = 13.6 $\times$ $\left( {{1 \over {{1^2}}} - {1 \over {{z^2}}}} \right)$

= ${3 \over 4} \times 13.6\,\,\,$ eV

Energy required to remove a electron from n^th excited state of doubly ionized lithium,

E₂ = ${{13.6{z^2}} \over {{n^2}}}$ = ${{13.6 \times {3^2}} \over {{n^2}}}$ eV

This energy is provided by the photon when it strike with the lithium atom.

$\therefore$    E₁ $\ge$ E₂

$\Rightarrow$     ${3 \over 4} \times 13.6 \ge {{13.6 \times 9} \over {{n^2}}}$

$\Rightarrow$    n²    $\ge$   3 $\times$ 4

$\Rightarrow$   n    $\ge$ $\sqrt {12}$

$\Rightarrow$     n $\ge$ 3.5

$\therefore$   Least possible excited state = 4