[qquad [Psi_(1,0)=((1)/(sqrt(pi(a_(0))^(3/2))),e^(-r/2_(0))*" The "r],[" crom the nucleus is (Hint: "(d^(2)Psi)/(dr^(2))=0],[(3)/(2)a_(0)," 4) "2a_(0)]]

psi_(1s)=(1)/(sqrt(pia_(0)^(3//2)))e^((r )/(a_(0))) Find the distance r from nucleous at which e^(-) finding probability is max. [For H-atom] a_(0)rarrI Bohr radius

The Wave function (Psi) of 2s is given by: Psi_(2s)=(1)/(2sqrt2pi)(1/a_(0))^(1//2){2-(r)/a^(0)}e^(-r//2a_(0)) At r =r_(0) , radial node is formed . Thus for 2s ,r_(0), in terms of a_(0) is-

[" The Schrodinger wave equation "],[" for hydrogen atom is: "],[qquad [psi_(2s)=(1)/(4sqrt(2 pi))((1)/(a_(0)))^(3/2)(2-(r_(0))/(a_(0)))e^((-r_(0))/(a_(0))),],[" where "a_(0)" is Bohr's radius.If the "],[" radial node is "2s" be at "r_(0)," then the "],[" value of "(r_(0))/(a_(0))" is "]]

Consider psi (wave function) of 2s atomic orbital of H-atom is- psi_(2s)=(1)/(4sqrt(2pia_(0)^(3//2)))[2-(r )/(a_(0))]e^.(r )/(2a_(0) Find distance of radial node from nucleous in terms of a_(0)

Consider psi (wave function) of 2s atomic orbital of H-atom is- psi_(2s)=(1)/(4sqrt(2pia_(0)^(3//2)))[2-(r )/(a_(0))]e^.(r )/(2a_(0) Find distance of radial node from nucleous in terms of a_(0)

Consider psi (wave function) of 2s atomic orbital of H-atom is- psi_(2s)=(1)/(4sqrt(2pia_(0)^(3//2)))[2-(r )/(a_(0))]e^.(r )/(2a_(0) Find distance of radial node from nucleous in terms of a_(0)

The Schrodinger wave equation for the 2s electron of a hydrogen atom is, psi_(2s)=(1)/(4sqrt(2pi))[(1)/(a_(0))]^(3//2)xx[2-(r)/(a_(0))]xxe^(-r//2a_(0)) There is a radial node at the point, where r=r_(0) find the relation between r_(0) and a_(0) .

[" The distance of spherical nodes "],[" from nucleus for the given orbital "],[" are "],[qquad [psi_(" radial ")=(1)/(9sqrt(2))((Z)/(a_(0)))^(3/2)[(sigma^(2)-4 sigma+3)]exp(-sigma/2],[" where "a_(0)&Z" are the constants and "],[sigma=(2Zr)/(a_(0))]]