For the wave function
`Psi= (sqrt2)/(81sqrtpi a_(0)^(3//2))[6 - (r)/(a_(0))](r)/(a_(0)) xx e^(-r//3a_(0)) "sin"theta dot"cos"phi`
Identify the orbital .

For 3s orbital of hydrogen atom, the normalised wave function is Psi_(3s)=(1)/((81)sqrt(3pi))((1)/(a_(o)))^(3//2)[27-(18r)/(a_(o))+(2r^(2))/(a_(o)^(2))]e^((-r)/(3a_(o))) If distance between the radial nodes is d, calculate rthe value of (d)/(1.73a_(o))

The wave function of 3s electron is given by Psi_(3s)=1/(81sqrt(3)prod)(1/a_(0))^(3//2)[27-18(r/a_(0))+2(r/a_(0))^(2)]e^(-r//3a_(0)) It has a node at r=r_(0) , Find out the relation between r_(0) and a_(0)

The Schrodinger wave equation for hydrogen atom is Psi_(2s) = (1)/(4sqrt(2pi)) ((1)/(a_(0)))^(3//2) (2 - (r)/(a_(0))) e^(-r//a_(0)) , where a_(0) is Bohr's radius . If the radial node in 2s be at r_(0) , then r_(0) would be equal to :

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)

a.The schrodinger wave equation for hydrogen atom is psi_(2s) = (1)/(4sqrt(2pi)) ((1)/(a_(0)))^((3)/(2)) (2 - (r_(0))/(a_(0)))e^((-(r )/(a)) When a_(0) is Bohr's radius Let the radial node in 2s be n at Then find r_(0) in terms of a_(0) b. A base ball having mass 100 g moves with velocity 100 m s^(-1) .Find the value of teh wavelength of teh base ball

The Schrodinger wave equation for hydrogen atom is psi_(2s) =1/(4sqrt(2pi)) (1/(a_(0)))^(3//2)(2-r/(a_(0)))e^(-t//a_(0)) where a_0 is Bohr's radius. If the radial node in 2 s be at r_0 , then r_0 would be equal to

The wave function of 2s electron is given by W_(2s) = (1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-1 a0) It has a node at r = r_(p) .Find the radiation between r_(p) and a