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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 $r = a_{0}$ ?

A

e

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B

e^{2}

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C

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

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D

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.

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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}})}

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}

?

Q. The first orbital of H is represented by

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

a_{0}

is Bohr's radius. The probability of finding the electron at a distance r from the nucleus in the region dV is?

Q. The first orbital of H is represented by:

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

a_{0}

is Bohr orbit. The probability of finding the electron at a distance 'r' from the nucleus is:

1 s

orbital of Hydrogen atom radial wave function is given as:

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

a_{0} = 0.529 A \circ)

The ratio of radial probability density of finding electron at

r = a_{o}

to the radial probability density of finding electron at the nucleus is given as (x.e

^{- y}

). Calculate the value of (x + y)

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

Above relation is wave function of hydrogen atom in ground state (

r =

distance of electron from nucleus and

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

1 s

orbital of hydrogen, the maximum probability of finding an electron is at

r =

:

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Shapes of Orbitals and Nodes

Standard XI Chemistry

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