Question

Bohr’s theory is applicable to:
$\left( A \right)$ $He$
$\left( B \right)$ $H{e^{2 + }}$
$\left( C \right)$ $L{i^{2 + }}$
$\left( D \right)$ $Li$

Answer 1

Hint: We know that Bohr proposed the mode of atoms on the basis of planck's quantum theory. This theory is primarily applicable for hydrogen or hydro like atoms. Electrons should move around the nucleus but only in prescribed orbitals.

Complete step by step answer:
First, we have to count the number of electrons in all the given species.
In $\left( {He} \right)$ , we know that the number of electrons in helium is $2$ but the Bohr’s theory is applicable only for one electron species. That’s why Bohr’s theory is not applicable for helium $\left( {He} \right)$ atoms.
In $\left( {H{e^{2 + }}} \right)$ , we know that the number of electron in helium is $2$ but $ + 2$ indicates that loss of two electrons to become helium ion$\left( {H{e^{2 + }}} \right)$, Hence there is no election and we know that Bohr’s theory is applicable one electron species. That’s why Bohr’s theory is not applicable for helium ion $\left( {H{e^{2 + }}} \right)$ .
In $\left( {L{i^{2 + }}} \right)$ , we know that the number of electron in lithium is $3$ but $ + 2$ indicates that loss of two electrons to become helium ion$\left( {L{i^{2 + }}} \right)$, Hence there is one election left and we know that Bohr’s theory is applicable one electron species. That’s why Bohr’s theory is applicable for lithium ion $\left( {L{i^{2 + }}} \right)$ .
In $\left( {Li} \right)$ , we know that the number of electrons in lithium is $3$ but the Bohr’s theory is applicable only for one electron species. That’s why Bohr’s theory is not applicable for the lithium $\left( {Li} \right)$ atom.
Hence, the correct option is $\left( {L{i^{2 + }}} \right)$ .

Note:
$1.$ The electron revolves around the nucleus in a well defined circular path or fixed radius and energy. $2.$ The energy does not change of its own with time. $3.$ electron received a required amount of energy jumps to a higher stationary state.