JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 11)
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 :
2
3
4
5
Explanation
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 nth 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
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 nth 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
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