The radial wave function for an orbital in a hydrogen atom is- - 116-3 1a0 32-x-1x2-8x-12-e-x2where- x-2ra0- a0- radius of first Bohr-s orbit- The minimum and maximum position of radial nodes from the nucleus are-

At radial node, ψ=0 ∴ From given equation, x 1=0 and x^2 8x+12=0 When x 1=0 ⇒ x=1 i.e., 2ra_0=1; r=a_02 (Minimum) When x^2 8x+12=0 (x 6)(x 2)=0 i) x 2 = 0 x=2 2 ...

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Standard XI

Chemistry

Shapes of Orbitals and Nodes

The radial wa...

Question

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:

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 $x^{2} - 8 x + 12 = 0$
When $x - 1 = 0$
$\Rightarrow x = 1$
i.e., $\frac{2 r}{a_{0}} = 1; r = \frac{a_{0}}{2}$ (Minimum)
When $x^{2} - 8 x + 12 = 0$
$(x - 6) (x - 2) = 0$
i) $x - 2 = 0$
$x = 2$
$\frac{2 r}{a_{0}} = 2,$ i.e., $r = a_{0}$ (Middle value)
ii) $x - 6 = 0$
$x = 6$
$\frac{2 r}{a_{0}} = 6$
$r = 3 a_{0}$ (Maximum)

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Similar questions

Q. 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:

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. What is the ratio of maximum and the middle positions of the radial nodes for an hydrogen atom that has this radial wave equation:

ψ = k {(\frac{1}{a_{o}})}^{3 / 2} [(x - 1) (x^{2} - 8 x + 12)] e^{- x / 2}

x = \frac{2 r}{a_{o}}, a_{o}

is radius of the first Bohr orbit.

Q. 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)

Q. For 3s orbital of hydrogen atom, the normalised wave function is

Ψ_{3 s} = \frac{1}{(81) \sqrt{3 π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}}] e^{\frac{- r}{3 a_{0}}}

If distance between the radial nodes is d, the value of

\frac{d}{1.5 a_{0}}

is (two decimal places)(Take

\sqrt{108} = 10

)

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Shapes of Orbitals and Nodes

Standard XI Chemistry

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The radial wave function for an orbital in a hydrogen atom is: ψ=116√3(1a0)32[(x−1)(x2−8x+12)]e−x2 where, x=2ra0; a0= radius of first Bohr's orbit. The minimum and maximum position of radial nodes from the nucleus are:

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