Ask your own question, for FREE!
Biology 17 Online
OpenStudy (anonymous):

what relationship exists between the mass number of an element and the isotopes of that element

OpenStudy (frostbite):

Well if you think about the mass of the atom there is a kind of relationship: The mass of the atom is determined by how mutch the isotopes weighs and the relative frequency of appearance. So if we for an example look at carbon, there are 2 natural isotopes: \[m \left(\begin{matrix}12 \\ 6\end{matrix}C\right)=12,00000000 u\] \[m \left(\begin{matrix}13 \\ 6\end{matrix}C\right)=13,00335483 u \] the two carbon isotopes occurs with the relative frequencies of 98.89% and 1.11% so the calculation is going to look like this: \[m(C) = 0,9889 . 12,00000000 u + 0,0111 . 13,00335483 u = 12,011 u\] This is precisely the value of the atomic mass of naturally occurring carbon.

OpenStudy (frostbite):

that dot between the procent and the mass is a multiplication sign :P

OpenStudy (frostbite):

You are familiar with isotopes right?

OpenStudy (frostbite):

"Give the symbol and number of neutrons in one atom of: (to get "mass number", you must round the "atomic mass" to the nearest whole number) Show your calculations." and we do it for barium, bismuth and carbon.

OpenStudy (anonymous):

and others lol

OpenStudy (frostbite):

Some notation first: \[_{Z}^{A}E\] Here is: Z the number of protons. A the nukleon-number/nucleus-number (the number of both protons and nutrons) E the shorten of the atom name we are talking about. Out from the we may be able to make the following equation to determine the number of neutrons (N): \[A=N+Z <-> N=A-Z\] We asume that the atom found is in then form with the highest relative frequency of appearance and are then able to do the following: m(E)=m(N) + m(Z) + m(electron) but becuase m(electron)<< m(N) and m(e)<< m(Z) and we set m(N) and m(Z) =1 we can rewrite to the following: m(E)=m(N) + m(Z) = A

OpenStudy (anonymous):

nevermind! my friends in the class before me, said that what she put was incorrect so we cant do this one :P

OpenStudy (frostbite):

Lets see how it works: \[_{4}^{9}B\] \[N=9-4=5\] \[_{83}^{209}Bi\] \[N=209-83=126\] \[_{6}^{12}C\] \[N=12-6=6\]

OpenStudy (frostbite):

Damn shame becuase it is damn easy :P

OpenStudy (anonymous):

lol xD

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!