Question 304: Atoms & Nuclei

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 :

Accepted Answer

Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,
E
1
=  13.6 $$ 	imes $$ $$\left( {{1 \over {{1^2}}} - {1 \over {{z^2}}}} ight)$$
= $${3 \over 4} 	imes 13.6\,\,\,$$ eV
Energy required to remove a electron from n
th
excited state of doubly ionized lithium,
E
2
= $${{13.6{z^2}} \over {{n^2}}}$$ = $${{13.6 	imes {3^2}} \over {{n^2}}}$$ eV
This energy is provided by the photon when it strike with the lithium atom.
$$ 	herefore $$   E
1
$$ \ge $$ E
2
$$ \Rightarrow $$   $${3 \over 4} 	imes 13.6 \ge {{13.6 	imes 9} \over {{n^2}}}$$
$$ \Rightarrow $$   n
2
$$ \ge $$  3$$ 	imes $$4
$$ \Rightarrow $$  n   $$ \ge $$ $$\sqrt {12} $$
$$ \Rightarrow $$    n $$ \ge $$ 3.5
$$ 	herefore $$  Least possible excited state = 4

Answer

2

Answer

3

Answer

4

Answer

5

Atoms & Nuclei

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 :

Choose the correct answer: