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Schrodinger Equation

The wave func...

Question

The wave function for $1 s$ orbital of the 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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1 / e^{2}

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Zero

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Solution

The correct option is D Zero
For $1 s$ orbital, probability of finding the electron at the nucleus is zero.

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Similar questions

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 =

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}})}^{\frac{3}{2}} e^{\frac{- r}{a_{0}}}

a_{0} = 0.529 \circ A

r = a_{0}

(x . e^{- y})

(x + y)

(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}

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