Question

why electron spin cant be 5 and -5?

Answer 1

i think @heena can give an appropriate answer..

Answer 2

well i m also not sure y not coz its the topic of electromagnetism and i didnt studied yet but acc to me wid d knowledge of atomic structure i ll say -ve sign is ignored as -ve only show direction means opp. and we always use mod i guess whenever we calulate so dat further calculation can be easier like if u r gettin one sin is 5 den u can urself say it direction means -ve and 5 a dey are in same orbital

Answer 3

actually i just wanted to know why it cant be natural numbers insted of 1/2?

Answer 4

ok it cannot be becoz electron can move posibbly in two direction only either upward or either downward dats y it is written +1/2 means upward and -1/2 means downward ok

Answer 5

why dont we take any other positive number as upward and its negative as downward>//?/

Answer 6

like?

Answer 7

u knw how to calculate probability?

Answer 8

5 and -5 or 3 adn -3.here also one is negative and other is +ve

Answer 9

are u asking about that schrodinger 's equation?

Answer 10

no not any equation ok come to maths coz its probability either upward or downward means two possible drection so total direction=2
now
total upward movemnt=1/2 and same for downward movement=1/2 no u ll get confuse so mention it dat -ve means downward and +ve means upward

Answer 11

actually why is 1/2 selected for representing electron spin?/

Answer 12

ok i knw i m too bad to xplain math topic and the reason is probability wait @sarkar come here right now

Answer 13

ok i m giving a last try as math is not my subject still let ssee u have coin and u tossed wat willl be the possible outcome?

Answer 14

u ll get tail or either head means two outcomes
now possible outcome of head=1/2
possible outcome of tail=1/2

Answer 15

actually the concept is clear.but is there any specific reason as to why 1/2 is chosen?spin of a photon is one .thats why i asked !!

Answer 16

it means u have only two direction and u h ave to put electron spin in diff two posible direction
dats y dey said 1/2

Answer 17

then why is the spin of photon 1

Answer 18

check these links,
they may halp you

http://www.newton.dep.anl.gov/askasci/chem99/chem99297.htm


http://answers.yahoo.com/question/index?qid=20110929035921AAadqTy

Answer 19

Indeed it does...and its interpretation is not trivial at all. A short answer is the following; atomic orbitals have integer amounts of angular momentum, right? An s orbital has l = 0, a p orbital has l = 1, etc...what's more, if you put a 2p electron into a magnetic field, you see a separation of the states in energy; these correspond to the "magnetic" quantum number, m_l, having values 1, 0, -1... these numbers represent how much of the orbital momentum can be projected along the z axis; if m_l = 1, then the projection = +h; if m_l = -1, the projection = -h; if zero, the projection = 0.
As it happens, if you turn up the strength of the magnetic field, you start to see each of the m_l states split into two states; each of which corresponds to an electron having m_s = 1/2 or m_s = -1/2...thus, if m_s = 1/2 there is an additional component of angular momentum along the z axis = +h/2; if m_s = -1/2 there's an additional component = -h/2.
So, the electron behaves as if it has "built-in" angular momentum which can add to, or subtract from, the orbital angular momentum...and it adds/subtracts half as much angular momentum projection as orbital angular momentum does. Whew!! I hope that helped. Some short answer!!!

Answer 20

I am at a loss of words,@heena

Answer 21

Spin actually has nothing to do with actual spinning, or rotation. It's a relavistic effect (like all magnetic effects), and it just so happens the math works out beautifully if you treat it AS IF it was angular momentum with half-integer quantum numbers, which, roughly speaking, means you have a half-integer number of de Broglie wavelengths along the circumference of one orbit. (Since that would mean a "wavefunction" would not satisfy boundary conditions of continuity, you can see immediately you can't construct any kind of genuine rotational wavefunction for spin -- which tells you it isn't, really, anything to do with rotation or spin at all.)