For 3s-orbital hydrogen atom, the normalized wave function isψ3s=1813π(1a0)3/2[27−18ra0+2r2a02]e−r/3a0 if distance between the radial nodes is d, calculate the value of d1.73a0

ψ3s2=0=1813π(1a0)3/2[27−18ra0+2r2a02]e−r/3a0 27−18ra0+2r2a02=0⇒r=18a0±10.39a04 Distance between to radial node d1.73a0=2×10.39a04×1.73a0 d=3

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$ψ_{3 s}^{2} = 0 = \frac{1}{81 \sqrt{3 π}} {(\frac{1}{a_{0}})}^{3 / 2} [27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{{a_{0}}^{2}}] e^{- r / 3 a_{0}} 27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{{a_{0}}^{2}} = 0 \Rightarrow r = \frac{18 a_{0} \pm 10.39 a_{0}}{4} D i s \tan c e b e t w e e n t o r a d i a l n o d e \frac{d}{1.73 a_{0}} = \frac{2 \times 10.39 a_{0}}{4 \times 1.73 a_{0}} d = 3$