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Shapes of Orbitals and Nodes

The wave func...

Question

The wave function for the $1 s$ orbital of a hydrogen atom is given by:
$Ψ_{1 s} = \frac{π}{\sqrt{2}} e^{(\frac{- r}{a_{0}})}$
where, $a_{0}$ = Radius of first Bohr's orbit
$r$ = Distance from the nucleus (probability of finding the electron varies with respect to it)

What will be the ratio of probabilities of finding the electron at the nucleus to the first Bohr's orbit at $r = a_{0}$ ?

e

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e^{2}

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\frac{1}{e^{2}}

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Zero

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Solution

The correct option is D Zero
For a $1 s$ or for that matter, any orbital electron, the probability of finding the electron at the nucleus is zero. Therefore, the ratio of probabilities of finding the electron at the nucleus to the first Bohr's orbit $a_{0}$ is zero since probability is zero at the nucleus.

Suggest Corrections

Similar questions

a_{0}

ψ = \frac{1}{\sqrt{π}} (\frac{1}{a_{0}})^{\frac{3}{2}} e^{\frac{- r}{a_{0}}},

a_{0}

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

a_{0}

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

a_{0}

1 s

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

a_{0} = 0.529 A \circ)

r = a_{o}

^{- y}

1 s

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

a_{0} = 0.529 \circ A

r = a_{0}

(x . e^{- y})

(x + y)

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