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Schrodinger Equation

The wave func...

Question

The wave function for $1 s$ orbital of hydrogen atom is given by:
$ψ_{1 s} = \frac{π}{\sqrt{2}} e^{- r / a_{0}}$
where, $a_{0} =$ Radius of first Bohr 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 electrons at the nucleus to first Bohr's orbit $a_{0}$ ?

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 any orbital, probability of finding the electron at the nucleus is zero. So ratio will also be zero.

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Similar questions

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

a_{0}

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

a_{0}

a_{0}

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

a_{0}

ψ = \frac{1}{4 a_{0} \sqrt{2 π a_{0}}} (2 - \frac{r}{a_{0}}) e^{\frac{- r}{2 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 =

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