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Bohr's Model of a Hydrogen Atom

What are the ...

Question

What are the average distance and the most probable distance of an electron from the nucleus in the 1s orbital of a hydrogen atom? $(a_{0}$ =the radius of the first Bohr orbit)

1.5 a_{0}

and

a_{0}

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a_{0}

and

5 a_{0}

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1.5 a_{0}

and

0.5 a_{0}

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a_{0}

and

0.5 a_{0}

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Solution

The correct option is B $1.5 a_{0}$ and $a_{0}$
The average distance and the most probable distance of an electron from the nucleus in the 1s orbital of a hydrogen atom are equal to $1.5 a_{0}$ and $a_{0}$ (Most probable distance is the distance where electron is highest probable so it is the radius of that orbit $a_{0}$ ). The value of $1.5 a_{0}$ is $0.529 ˚ A$ . Hence, the average distance and the most probable distance of an electron from the nucleus in the 1s orbital of a hydrogen atom are equal to $0.794 ˚ A$ and $0.529 ˚ A$ respectively.

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Similar questions

1 s

Ψ_{1 s} = \frac{π}{\sqrt{2}} e^{- r / a_{0}}

a_{0} =

r =

a_{0}

Ψ = \frac{1}{\sqrt{π}} {[\frac{1}{a_{0}}]}^{3 / 2} e^{- r / a_{0}}

a_{0}

Ψ_{1 s} = [\frac{1}{\sqrt{π a_{o}^{3}}}] e^{- r / a_{0}}

r =

a_{0} = 0.529 \times 10^{- 8} c m

1 s

r =

0.529 A^{\circ}

Ψ = \sqrt{(\frac{Z^{3}}{π a_{0}^{3}})} e^{- Z r / a_{0}}

a_{0}

Ψ = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} e^{- π / a_{0}}

a_{0}

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