The schrodinger wave equation for hydrogen atom is `Psi_2 = (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` would be equal to
A
`(a_(0))/(2)`
B
`2a_(0)`
C
`sqrt(2)a_(0)`
D
`(a_(0))/(sqrt(2))`
Text Solution
Verified by Experts
Topper's Solved these Questions
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise Level- 1|1 Videos
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise level 2|1 Videos
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 2|1 Videos
Similar Questions
Explore conceptually related problems
The radial distribution function [P(r)] is use 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^(-2Ze//a_(0)) . Then which of the following statements is/are correct
The radius of first Bohr.s orbit for hydrogen is 0.53 Å. The radius of third Bohr.s orbit would be
If (1)/(sqrt(4x+1)){(1+(sqrt(4x+1))/(2))^n-(1-(sqrt(4x+1))/(2))^n}=a_0+a_1 x+...+a_5 x^5 , then n=
From quantization of angular momentum, one gets for hydrogen atom. the radius of the n^(th) orbit as r_(n) = (n^(2))/(m_(e)) ((h)/(2 pi))^(2) ((4 pi^(2) epsilon_(0))/(e^(2)) ) for a hydrogen like atom of atomic number 'Z'.