For an electr...

Question

For an electron in a hydrogen atom, the wave function $Ψ$ is proportional to exp, where $a_{0}$ is the Bohrs radius. What is the ratio of the probability of finding the electron at the nucleus to the probability of finding it at $a_{0}$ ?

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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
Electron cannot be found at nucleus.
$∴$ Probability finding electron at nucleus is zero.

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a_{0}

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

a_{0}

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a_{0} =

r =

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

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

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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})

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

a_{0}