Find the distance at which probability of finding electron is maximum for `1s` orbital in a He atom. The wave function orbital given as.
`psi_(1s)=(4)/(a_(0)^(3//2))e^((2r)/(a_(0))`

Probability of finding the electron psi^(2) of 's' orbital does not depend upon

The radius of maximum probability for 1s orbital of H atom is

One of the major requirement in atomic structure is determination of location of electron inside an atom. The wave mechanical model establishes this in accordance with Heisenberg's uncertainity principle through the concept of orbitals. The orbitals are defined as that '3D' space in which probability of finding electron is maximum and are represented by wave functions Psi_(n,l,m) where n,l and m are quantum number. The variation of Psi is analysed in terms of polar coordinates and hence Psi = f( r, 0 , phi) where 'r' represents radius vector and 0 and phi represents angle (/_) Which the radius vector with x-axis respectively. The expressions of Psi_(r,0, phi) are often given in terms of sigma instead of r where sigma=(2Zr)/(nalpha_(0)) and Z = atomic number and n= shell number. Which of the following wave function cannot represent an atomic orbital of H-atom?

One of the major requirement in atomic structure is determination of location of electron inside an atom. The wave mechanical model establishes this in accordance with Heisenberg's uncertainity principle through the concept of orbitals. The orbitals are defined as that '3D' space in which probability of finding electron is maximum and are represented by wave functions Psi_(n,l,m) where n,l and m are quantum number. The variation of Psi is analysed in terms of polar coordinates and hence Psi = f( r, 0 , phi) where 'r' represents radius vector and 0 and phi represents angle (/_) Which the radius vector with x-axis respectively. The expressions of Psi_(r,0, phi) are often given in terms of sigma instead of r where sigma=(2Zr)/(nalpha_(0)) and Z = atomic number and n= shell number. If an orbital is represented as: Psi_(r,0,phi)=2/3((1)/(3alpha_(0)))^(3//2).(sigma-1)(sigma^(2)-8sigma+12) .sigma^(-sigma//2).cos0 belong to which orbital?

The radial distribution functions [P(r)] is used to determine the most probable radius, which is used to find the electron in a given orbital (dP(r))/(dr) for 1s -orbital of hydrogen like atom having atomic number Z , is (dP)/(dr)=(4Z^(3))/(a_(0)^(3))(2r-(2Zr^(2))/(a_(0)))e^(-2Zr//a_(0)) :

The Schrodinger wave equation for H-atom is nabla^(2) Psi = (8pi^(2)m)/(h^(2)) (E-V) Psi = 0 Where nabla^(2) = (del^(2))/(delx^(2)) +(del^(2))/(dely^(2)) +(del^(2))/(delz^(2)) E = Total energy and V=potential energy wave function Psi_(((r, theta,phi)))R_((r))Theta_((theta))Phi_((phi)) R is radial wave function which is function of ''r'' only, where r is the distance from nucleus. Theta and Phi are angular wave function. R^(2) is known as radial probability density and 4pir^(2)R^(2)dr is known as radial probability function i.e., the probability of finding the electron is spherical shell of thickness dr. Number of radial node =n -l - 1 Number of angular node = l For hydrogen atom, wave function for 1s and 2s-orbitals are: Psi_(1s) = sqrt((1)/(pia_(0)^(a)))e^(-z_(r)//a_(0)) Psi_(2s) = ((Z)/(2a_(0)))^(½) (1-(Zr)/(a_(0)))e^(-(Zr)/(a_(0))) The plot of radial probability function 4pir^(2)R^(2) aganist r will be: Answer the following questions: The following graph is plotted for ns-orbitals The value of 'n' will be:

The Schrodinger wave equation for H-atom is nabla^(2) Psi = (8pi^(2)m)/(h^(2)) (E-V) Psi = 0 Where nabla^(2) = (del^(2))/(delx^(2)) +(del^(2))/(dely^(2)) +(del^(2))/(delz^(2)) E = Total energy and V=potential energy wave function Psi_(((r, theta,phi)))R_((r))Theta_((theta))Phi_((phi)) R is radial wave function which is function of ''r'' only, where r is the distance from nucleus. Theta and Phi are angular wave function. R^(2) is known as radial probability density and 4pir^(2)R^(2)dr is known as radial probability function i.e., the probability of finding the electron is spherical shell of thickness dr. Number of radial node =n -l - 1 Number of angular node = l For hydrogen atom, wave function for 1s and 2s-orbitals are: Psi_(1s) = sqrt((1)/(pia_(0)^(a)))e^(-z_(r)//a_(0)) Psi_(2s) = ((Z)/(2a_(0)))^(½) (1-(Zr)/(a_(0)))e^(-(Zr)/(a_(0))) The plot of radial probability function 4pir^(2)R^(2) aganist r will be: Answer the following questions: The value of radius 'r' for 2s atomic orbital of H-atom at which the radial node will exist may be given as:

The Schrodinger wave equation for H-atom is nabla^(2) Psi = (8pi^(2)m)/(h^(2)) (E-V) Psi = 0 Where nabla^(2) = (del^(2))/(delx^(2)) +(del^(2))/(dely^(2)) +(del^(2))/(delz^(2)) E = Total energy and V=potential energy wave function Psi_(((r, theta,phi)))R_((r))Theta_((theta))Phi_((phi)) R is radial wave function which is function of ''r'' only, where r is the distance from nucleus. Theta and Phi are angular wave function. R^(2) is known as radial probability density and 4pir^(2)R^(2)dr is known as radial probability function i.e., the probability of finding the electron is spherical shell of thickness dr. Number of radial node =n -l - 1 Number of angular node = l For hydrogen atom, wave function for 1s and 2s-orbitals are: Psi_(1s) = sqrt((1)/(pia_(0)^(a)))e^(-z_(r)//a_(0)) Psi_(2s) = ((Z)/(2a_(0)))^(½) (1-(Zr)/(a_(0)))e^(-(Zr)/(a_(0))) The plot of radial probability function 4pir^(2)R^(2) aganist r will be: Answer the following questions: What will be number of angular nodes and spherical nodes for 4f atomic orbitals respectively.

A wave function for an atomic orbital is given as Psi_(2,1,0) Recogine the orbital