For an electron in a hydrogen atom, what is the ratio of probability density of finding the electron at the nucleus to the probability density of finding it at $a_{0}$ ,the wave function is $ψ = \frac{1}{\sqrt{π}} (\frac{1}{a_{0}})^{\frac{3}{2}} e^{\frac{- r}{a_{0}}},$ where $a_{0}$ is Bohr’s radius

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 B $e^{2}$
probability density of finding electron $= ψ^{2}$
So, Ratio $= \frac{ψ_{r = 0}^{2}}{ψ_{r = a_{0}}^{2}} = \frac{1}{(e^{- 1})^{2}} = e^{2}$

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Similar questions

Q. 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}

For the $1 s$ orbital of the Hydrogen atom, the radial wave function is given as:
$R (r) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{\frac{- r}{a_{0}}}$ (Where $a_{0} = 0.529 \circ A$ )

The ratio of radial probability density of finding an electron at $r = a_{0}$ to the radial probability density of finding an electron at the nucleus is given as $(x . e^{- y})$ .
Calculate the value of $(x + y)$ .