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Question Number 16137 by Tinkutara last updated on 18/Jun/17

For 2s orbital Ψ_r  = (1/(√8))((z/a_0 ))^(3/2) (2 − ((zr)/a_0 ))e^(−((zr)/(2a_0 )))   then, hydrogen radial node will be at  the distance of  (1) a_0   (2) 2a_0   (3) (a_0 /2)  (4) (a_0 /3)

$$\mathrm{For}\:\mathrm{2}{s}\:\mathrm{orbital}\:\Psi_{\mathrm{r}} \:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{8}}}\left(\frac{\mathrm{z}}{\mathrm{a}_{\mathrm{0}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{2}\:−\:\frac{\mathrm{zr}}{\mathrm{a}_{\mathrm{0}} }\right)\mathrm{e}^{−\frac{\mathrm{zr}}{\mathrm{2a}_{\mathrm{0}} }} \\ $$$$\mathrm{then},\:\mathrm{hydrogen}\:\mathrm{radial}\:\mathrm{node}\:\mathrm{will}\:\mathrm{be}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{a}_{\mathrm{0}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2a}_{\mathrm{0}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{3}} \\ $$

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