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Standard XII

Chemistry

Visualising Orbitals

The radial wa...

Question

The radial wave equation for hydrogen atom is
$Ψ = \frac{1}{16 \sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} [(x - 1) (x^{2} - 8 x + 12)] e^{- x / 2}$
where, $x = 2 r / a_{0}$ ; $a_{0} =$ radius of first Bohr orbit.

The minimum and maximum position of radial nodes from nucleus are:

a_{0}, 3 a_{0}

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\frac{a_{0}}{2}, 3 a_{0}

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\frac{a_{0}}{2}, a_{0}

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\frac{a_{0}}{2}, 4 a_{0}

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Solution

The correct option is B $\frac{a_{0}}{2}, 3 a_{0}$
At radial node, $Ψ = 0$

$∴$ From given equation,
$x - 1 = 0$ and $x^{2} - 8 x + 12 = 0$

$x - 1 = 0 \Rightarrow x = 1$

i.e., $\frac{2 r}{a_{0}} = 1$ ; $r = \frac{a_{0}}{2}$ (Minimum)

$x^{2} - 8 x + 12 = 0$

$(x - 6) (x - 2) = 0$
when $x - 2 = 0$
$x = 2$

$\frac{2 r}{a_{0}} = 2$ , i.e., $r = a_{0}$ (Middle value)

when $x - 6 = 0$
$x = 6$

$\frac{2 r}{a_{0}} = 6$

$r = 3 a_{0}$ (Maximum)

Hence, option B is correct.

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Similar questions

Q. The radial wave function for an orbital in a hydrogen atom is:

ψ = \frac{1}{16 \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [(x - 1) (x^{2} - 8 x + 12)] e^{- \frac{x}{2}}

where,

x = \frac{2 r}{a_{0}}; a_{0} =

radius of first Bohr's orbit. The minimum and maximum position of radial nodes from the nucleus are:

Q. The Schrodinger wave function for Hydrogen atom of 4s orbital is given by:

Ψ_{4 s} = \frac{1}{16 \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [(σ - 1) (σ^{2} - 8 σ + 12)] e^{- σ / 2}

where

a_{0}

1^{s t}

Bohr radius and

σ = \frac{2 r}{a_{o}}

.

The distance from the nucleus where there is no radial node will be:

Q. The Schrondinger equation for hydrogen atom is

ψ = \frac{1}{4 \sqrt{2 π}} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r_{0}}{a_{0}}) e^{-} \frac{r_{0}}{a_{0}}

, where

a_{0}

is Bohr's radius. If the radial node is

2 s

be at

r_{0}

.
Then,

r_{0}

in terms of

a_{0}

is given as

x a_{0} = r_{0}

, value of

x

Q. The Schrondinger equation for hydrogen atom is

ψ = \frac{1}{4 \sqrt{2 π}} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r_{0}}{a_{0}}) e^{-} \frac{r_{0}}{a_{0}}

, where

a_{0}

is Bohr's radius. If the radial node is

2 s

be at

r_{0}

.
Then,

r_{0}

in terms of

a_{0}

is given as

x a_{0} = r_{0}

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Q. For the

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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

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to the radial probability density of finding an electron at the nucleus is given as

(x . e^{- y})

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(x + y)

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The radial wave equation for hydrogen atom is Ψ=116√π(1a0)3/2[(x−1)(x2−8x+12)]e−x/2where, x=2r/a0; a0= radius of first Bohr orbit.The minimum and maximum position of radial nodes from nucleus are:

The correct option is B a02,3a0At radial node, Ψ=0∴ From given equation,x−1=0 and x2−8x+12=0x−1=0⇒x=1i.e., 2ra0=1; r=a02 (Minimum)x2−8x+12=0(x−6)(x−2)=0when x−2=0 x=2 2ra0=2, i.e., r=a0 (Middle value)when x−6=0 x=6 2ra0=6 r=3a0 (Maximum)Hence, option B is correct.

The radial wave equation for hydrogen atom is
$Ψ = \frac{1}{16 \sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} [(x - 1) (x^{2} - 8 x + 12)] e^{- x / 2}$
where, $x = 2 r / a_{0}$ ; $a_{0} =$ radius of first Bohr orbit.

The minimum and maximum position of radial nodes from nucleus are: