
Calculate the radius of orbit of \[{\text{H}}{{\text{e}}^{\text{ + }}}\] ion in the first orbit.
A.\[0.2645\]\[{{\text{A}}^{\text{o}}}\]
B.\[26.45\]\[{{\text{A}}^{\text{o}}}\]
C.\[0.2465 \times {10^{ - 12}}\]\[{{\text{A}}^{\text{o}}}\]
D.None of these
Answer
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Hint: We should know that the radius of the orbit of \[H{e^ + }\] refers to the Bohr radius. In a hydrogen atom in its ground state, the Bohr radius is the most likely distance between the nucleus and the electron. It was named after Niels Bohr because of its function in the Bohr atom model. We should know that in the Bohr model for atomic structure, put forward by Niels Bohr in 1913, electrons orbit a central nucleus under electrostatic attraction.
Complete answer:
The following formula can be used to calculate the radius of Bohr's orbit in hydrogen and hydrogen-like species,
\[radius of orbit = r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}}}{\text{ }} \times \dfrac{1}{Z} = 0.529 \times \dfrac{{{n^2}}}{Z}{A^o}\]
Where n is the principal quantum number of orbit and
Z is the atomic number.
For the first orbit of \[H{e^ + }\] ion, the principal quantum number is $1$.
Therefore, $n = 1$
An atomic number for \[H{e^ + }\] ion is $2$.
Therefore, $z = 2$
Then, the radius will be given by,
\[r = 0.529 \times \dfrac{{{1^2}}}{2}\]
On simplification we get,
$r = 0.2645{A^o}$
Thus, the correct option is A.
Note:
We need to know that the electron probability cloud obeying the Schrodinger equation has superseded the Bohr model of the atom, which is further complicated by spin and quantum vacuum effects to produce fine and hyperfine structure. The Bohr radius specifies the radius with the largest radial probability density, not the predicted radial distance, which is a critical distinction. Because of the lengthy tail of the radial wave function, the anticipated radial distance is $1.5$ times the Bohr radius.
Additional information: Another major distinction is that the greatest probability density in three-dimensional space occurs at the nucleus' location, not at the Bohr radius, although the radial probability density peaks at the Bohr radius, i.e. when plotting the probability distribution in its radial dependency.
Complete answer:
The following formula can be used to calculate the radius of Bohr's orbit in hydrogen and hydrogen-like species,
\[radius of orbit = r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}}}{\text{ }} \times \dfrac{1}{Z} = 0.529 \times \dfrac{{{n^2}}}{Z}{A^o}\]
Where n is the principal quantum number of orbit and
Z is the atomic number.
For the first orbit of \[H{e^ + }\] ion, the principal quantum number is $1$.
Therefore, $n = 1$
An atomic number for \[H{e^ + }\] ion is $2$.
Therefore, $z = 2$
Then, the radius will be given by,
\[r = 0.529 \times \dfrac{{{1^2}}}{2}\]
On simplification we get,
$r = 0.2645{A^o}$
Thus, the correct option is A.
Note:
We need to know that the electron probability cloud obeying the Schrodinger equation has superseded the Bohr model of the atom, which is further complicated by spin and quantum vacuum effects to produce fine and hyperfine structure. The Bohr radius specifies the radius with the largest radial probability density, not the predicted radial distance, which is a critical distinction. Because of the lengthy tail of the radial wave function, the anticipated radial distance is $1.5$ times the Bohr radius.
Additional information: Another major distinction is that the greatest probability density in three-dimensional space occurs at the nucleus' location, not at the Bohr radius, although the radial probability density peaks at the Bohr radius, i.e. when plotting the probability distribution in its radial dependency.
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