`n(phi)`=

A

n

B

`phi`

C

0

D

9

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • SETS

    BEYOND PUBLICATION|Exercise EXERCISE|247 Videos

  • REAL NUMBERS

    BEYOND PUBLICATION|Exercise |1 Videos

  • SIMILAR TRIANGLES

    BEYOND PUBLICATION|Exercise EXERCISE|143 Videos

Similar Questions

Explore conceptually related problems

If phi(x) is polynomial function and phi(x)gtphi(x)AAxge1andphi(1)=0 ,then

If (1+tantheta)(1+tanphi)=2 then theta+phi =

{:(" Column- I "," Column-II "),("A) If " ((sin theta)/(sin phi))^(2) = (tan theta)/(tan phi)=3 "then the value of " theta and phi " are ","(p) Infinite number of solutions "),("B) The number of solutions of " (sin ^(2) x + cos ^(4) x)/(cos^(2) x + sin^(4) x ) = 1 ,"q) " theta = n pi pm (pi)/(3)"," phi = n pi pm (pi)/(6) ),("(c) Solution of " cot 2 theta = cot ^(2) theta - tan ^(2) theta ,"r) " n pi pm (pi)/(4)):}

sin(90-phi) = ……………..

State which of the following statement are true. (i) {}=phi (ii) phi=0 (iii) 0={0}

IF theta-phi=pi/2 , then show that [{:(cos^2theta,costhetasintheta),(costhetasintheta,sin^2theta):}][{:(cos^2phi,cosphisinphi),(cosphisinphi,sin^2phi):}]=O

When Schrodinger wave equation in polar coordinates is solved the solution for Phi is of the form Psi (r, theta , phi)= R(r) , Y(theta , phi) . Here R(r) is radial part of wave function and Y(theta, phi) is angular part of the wave function. The region or space where probability of finding electron is zero is called nodal surface. If the probability of finding electron is zero then Psi^2 (r, theta, phi)=0 implies Psi (r, theta, phi)=0 If the radial wave function is equal to zero we get radial node and if angular part is equal to zero we get angular nodes. Total no. of nodes for any orbital = n - 1. Where ‘n’ is principal quantum number. Number of radial nodes for 4f orbital