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Standard XII

Chemistry

Heisenberg's Uncertainty Principle

For the 1s or...

Question

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 $(x + y)$ .

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Solution

$\frac{Radial probability density at r = a_{0}}{Radial probability density at r = 0} = \frac{R^{2} (a_{0})}{R^{2} (0)}$
For 1s orbital: $R_{(r)} = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{\frac{- r}{a_{0}}}$
Hence,
$\frac{R^{2} (a_{0})}{R^{2} (0)} = \frac{(\frac{1}{π a_{0}^{3}}) e^{\frac{- 2 r}{a_{0}}}}{(\frac{1}{π a_{0}^{3}}) e^{0}} = e^{- 2}$
Hence, according to $x . e^{- y}$ given in question.
Here, $x = 1, y = 2$ and $(X + Y) = 3$

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Similar questions

Q. For

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}}

where

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)

Q. 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

Q. The wave function for

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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}

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 (

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distance of electron from nucleus and

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

). For

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Q. The first orbital of H is represented by

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

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

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Radial probability density at r=a0Radial probability density atr=0=R2(a0)R2(0)For 1s orbital: R(r)=1√π(1a0)32e−ra0Hence, R2(a0)R2(0)=(1πa30)e−2ra0(1πa30)e0=e−2Hence, according to x.e−y given in question.Here, x=1, y=2 and (X+Y)=3

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