For 3s orbital of hydrogen atom- the normalised wave function is -3s - 181-3 - 1a0 32 - 27 - 18ra0 - 2r2a20 - e-r3a0 If distance between the radial nodes is d- the value of d1-5 a0 is two decimal placesTake -108 - 10

At Radial node the wave function becomes zero i.e. nbsp; Ψ_3s = 0 181 √(3 π) ( 1a_0 )^32 [ 27 18ra_0 + 2r^2a_0^2 ] e^ r3a_0 = 0 now since the exponential ...

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Byju's Answer

Standard XIII

Chemistry

Bohr's Model of a Hydrogen Atom

For 3s orbita...

Question

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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Solution

At Radial node the wave function becomes zero i.e.
$Ψ_{3 s} = 0$
$\frac{1}{81 \sqrt{3 π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}}] e^{- \frac{r}{3 a_{0}}} = 0$

now since the exponential part cannot be zero as then we have r = inifinity which is not possible. so the polynomial part will be zero.
$27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}} = 0$
$2 r^{2} - 18 a . r + 27 a_{0}^{2} = 0$
using formula of finding roots of quadratic equation :
$r = - \frac{- (- 18 a_{0}) + - \sqrt{(- 18 a_{0})^{2} - 4 \times 2 \times 27 a_{0}^{2}}}{2 \times 2}$

$r = \frac{18 a_{0} + - 10 a_{0}}{4}$

this will give us two roots of r (i.e two radial nodes) lets call them $r_{1}$ and $r_{2}$ so,
$r_{1} = \frac{18 a_{0} + 10 a_{0}}{4}$

$r_{2} = \frac{18 a_{0} - 10 a_{0}}{4}$
so distance between the nodes will be:
$r_{1} - r_{2} = d = 2 \times \frac{10 a_{0}}{4}$

$\frac{d}{1.5 a_{0}} = \frac{2 \times 10 a_{0}}{4 \times 1.5 a_{0}} = 3.33$

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For 3s orbital of hydrogen atom, the normalised wave function is Ψ3s=1(81)√3π(1a0)32[27−18ra0+2r2a20]e−r3a0 If distance between the radial nodes is d, the value of d1.5 a0 is (two decimal places)(Take √108=10 )