What is the correct relationship?
(A) ${{E}_{1}}ofH=1/2{{E}_{2}}ofH{{e}^{+}}=1/3{{E}_{3}}ofL{{i}^{2+}}=1/4{{E}_{4}}ofB{{e}^{3+}}$
(B) ${{E}_{1}}(H)={{E}_{2}}(H{{e}^{+}})={{E}_{3}}(L{{i}^{2+}})={{E}_{4}}(B{{e}^{3+}})$
(C) ${{E}_{1}}(H)=2{{E}_{2}}(H{{e}^{+}})=3{{E}_{3}}(L{{i}^{2+}})=4{{E}_{4}}(B{{e}^{3+}})$
(D) No relation
Answer
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Hint: Hydrogen atom is the simplest atom to apply Bohr model and Schrodinger equation. An atom in its normal state has Z protons and Z electrons in order to be neutral. Z is called the atomic number. From the Bohr’s equation, can calculate the energy of electron like single atoms H, $H{{e}^{+2}},L{{i}^{+3}} and,B{{e}^{+4}}$ .
Complete answer:
In the equation given, ${{E}_{n}}$ = energy of nth orbit
We know, hydrogen like single atom, energy of the nth orbit, \[{{E}_{n}}=-13.6X\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV\] -- (1)
Where, Z = atomic number of atoms, and n= number of orbits. This equation is known as energy of electrons in the Bohr orbit of the H atom.
Now, for H atom, the first energy level, ${{E}_{1}}$ will be, (from equation-1)
\[{{E}_{1}}=-13.6\dfrac{{{(1)}^{2}}}{{{(1)}^{2}}}=-13.6eV\]--- (2), where Z= atomic number of H =1 and energy level n =1
For $H{{e}^{+2}}$ atom, ${{E}_{2}}$ will be (from equation-1),
\[{{E}_{2}}=-13.6\dfrac{{{(2)}^{2}}}{{{(2)}^{2}}}=-13.6eV\]--- (3), where Z= 2 for He and n=2
For $L{{i}^{+3}}$ , ${{E}_{3}}$ will be (from equation-1),
\[{{E}_{3}}=-13.6\dfrac{{{(3)}^{2}}}{{{(3)}^{2}}}=-13.6eV\] -- (4), where Z=3 for Li and n=3
For $B{{e}^{+4}}$ , ${{E}_{4}}$ will be (from equation-1),
\[{{E}_{4}}=-13.6\dfrac{{{(4)}^{2}}}{{{(4)}^{2}}}=-13.6eV\] -- (5), where Z=4 for Be and n=3
From equation (2), (3), (4) and (5), the energy of electrons for a single electron atom is of equal value.
Hence, ${{E}_{1}}(H)={{E}_{2}}(H{{e}^{+}})={{E}_{3}}(L{{i}^{2+}})={{E}_{4}}(B{{e}^{3+}})$
So the correct answer is option B
Note:
An electron in an orbit is to ignore the interactions between electrons with each other and assume each electron is moving under the action of the nucleus as a point charge +Ze. Each electron has an independent potential energy and wave function.
Complete answer:
In the equation given, ${{E}_{n}}$ = energy of nth orbit
We know, hydrogen like single atom, energy of the nth orbit, \[{{E}_{n}}=-13.6X\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV\] -- (1)
Where, Z = atomic number of atoms, and n= number of orbits. This equation is known as energy of electrons in the Bohr orbit of the H atom.
Now, for H atom, the first energy level, ${{E}_{1}}$ will be, (from equation-1)
\[{{E}_{1}}=-13.6\dfrac{{{(1)}^{2}}}{{{(1)}^{2}}}=-13.6eV\]--- (2), where Z= atomic number of H =1 and energy level n =1
For $H{{e}^{+2}}$ atom, ${{E}_{2}}$ will be (from equation-1),
\[{{E}_{2}}=-13.6\dfrac{{{(2)}^{2}}}{{{(2)}^{2}}}=-13.6eV\]--- (3), where Z= 2 for He and n=2
For $L{{i}^{+3}}$ , ${{E}_{3}}$ will be (from equation-1),
\[{{E}_{3}}=-13.6\dfrac{{{(3)}^{2}}}{{{(3)}^{2}}}=-13.6eV\] -- (4), where Z=3 for Li and n=3
For $B{{e}^{+4}}$ , ${{E}_{4}}$ will be (from equation-1),
\[{{E}_{4}}=-13.6\dfrac{{{(4)}^{2}}}{{{(4)}^{2}}}=-13.6eV\] -- (5), where Z=4 for Be and n=3
From equation (2), (3), (4) and (5), the energy of electrons for a single electron atom is of equal value.
Hence, ${{E}_{1}}(H)={{E}_{2}}(H{{e}^{+}})={{E}_{3}}(L{{i}^{2+}})={{E}_{4}}(B{{e}^{3+}})$
So the correct answer is option B
Note:
An electron in an orbit is to ignore the interactions between electrons with each other and assume each electron is moving under the action of the nucleus as a point charge +Ze. Each electron has an independent potential energy and wave function.
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