There can only be 2 electrons in each single orbital, and they will be on opposite sides of the electron cloud (orbital).
Each planet in the solar system has a different orbital period, corresponding to the different sizes of their elliptical orbits.For the Earth, the present orbital period is 365.25636days. (rounded)
Four pairs of electrons, with one unpaired. This is the electronic configuration of fluorine
13
If the Sun were to suddenly grow by 50% in mass, at its present orbital speed, the Earth would not travel quickly enough around the sun to remain in orbit, and the Earth would fall into a spiral collision course with the Sun..In order for the the Earth to maintain its present orbital radius with a 50% more massive Sun, the Earth's linear orbital velocity would have to be 22% faster than it is now. A year would only be 298.2 days long (9.8 months), instead of 365.25 days..Another way of looking at it, in order to maintain its present orbital period of 365.25 days, the Earth would have to increase its linear orbital velocity by 14.5% and increase its orbital radius also by 14.5%..If it were 50% more massive, the Sun may also be considerably hotter, and therefore the Earth would be hotter, in which case the Earth would likely not be able to sustain life.
three nodes
There can be 10 electrons in the n=2 shell. Two can fit in the 1s orbital, two can fit in the 2s orbital, and six can fit in the 2p orbital.
vacant d orbital means an empty d orbital. For example, in nitrogen a d orbital is not allowed whereas in phosphorus a vacant d orbital is present.
The s orbital is present in all valid principal quantum number shells.The p orbital is present in n = 2 and higher.The d orbital is present in n = 3 and higher.The f orbital is present in n = 4 and higher.So the invalid ones are b (there are no 2d orbitals) and c (there are no 3f orbitals). 4s and 3p are perfectly legitimate.
n1= 25 n2= n1+1 n3= n1-1 n4=n1+2 n5=n1-2
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
The N1 is a rocket.
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
the value of the exponent n1
the value of the exponent n1
With greyhond