For 3s orbital of hydrogen atom, the normalized wave function is ψ3S=1813π1a03/227−18ra0+2r2a02e−r3a0 If distance between the radial nodes is d. Calculate the value of d1.73 a0

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For 3s orbital of hydrogen atom, the normalized wave function is $ψ_{3 S} = \frac{1}{81 \sqrt{3 π}} {(\frac{1}{a_{0}})}^{3 / 2} [27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}}] e^{\frac{- r}{3 a_{0}}}$
If distance between the radial nodes is d. Calculate the value of $\frac{d}{1.73 a_{0}}$

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answer is 3.

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Detailed Solution

$ψ^{2} = 0 \Rightarrow 27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}} = 0 2 r^{2} - 18 a_{0} r + 27 a_{0}^{2} = 0 r = \frac{18 a_{0} \pm 10.39 a_{0}}{4} s o l v i n g t w o v a l u e s a r e o b t a i n e d 3.76 a n d 1 d i s \tan c e b e t w e e n r a d i a l n o d e s (d) = 3.76 - 1 = 2.76 \frac{d}{1.730 a_{0}} = \frac{2.76}{1.73 \times 0.53} = 3$