Question

if phosphorus (P) has only one stabile isotope why is his atomic mass 30,9 and not exactly 31?

Answer 1

because they isotopes still exist, regardless of their instability

Answer 2

the*

Answer 3

although they're a small fraction, which is observed as the small change from 31 to 30.9 in molecular mass

Answer 4

everywhere i have looked it says that 31P is 100% present in nature, and that sytetic isotopes are not counted in because they have very short half life and can only exist in laboratory conditions.. is it posible that 30,9 is because of something else?

Answer 5

umm, it could be due to the mass lost due to the binding energy

Answer 6

ok, i think thats it :) can you explain me that with more details?

Answer 7

basically, in order for particles to stick together at the nucleus they need some to invest some energy, which equates to mass E=mc^2.. this energy can be liberated when they're separated (e.g.. nuclear fission)

Answer 8

thank you :) you've saved my day ;)

Answer 9

anytime (;

Answer 10

When we write the isotopes of an element, the integers we write are NOT the atomic masses, but what is called the "atomic mass number," equal to the number of protons and neutrons in the nuclei. So \[{}^{31}{\rm P}\] has 31 protons and neutrons -- but that does NOT mean its mass is 31 amu. In fact, an atom of P-31 weighs 30.9737620 amu, which is certainly close to 31, but not exactly.

You should bear in mind that in general atomic masses are close to, but not exactly equal to, atomic mass numbers, for two reasons:

(1) the mass of the proton and neutron are not exactly 1 amu. The mass of a free proton is 1.007276467 amu, and the mass of a free neutron is 1.008664916 amu.

(2) When protons and neutrons combine into a nucleus, a great deal of energy is released, and this results in a reduction of the mass of the nucleus below what the sum of the masses of free protons and neutrons would imply. For example, two free protons and two free neutrons weigh 4.031883 amu, but a nucleus of helium-4 weighs only 4.002603 amu. The "missing" 0.02928 amu represents the energy lost when a He-4 nucleus is formed from 2 free protons and 2 free neutrons. That's the amount of energy that must be supplied to break up the nucleus again.

This "binding energy" various in complex ways from nucleus to nucleus, so you can't easily say what it will be for a given isotope.