I have no idea how to do this...any ideas?
Are u sure with the numbers??U said 158,160 & 160.
Do you know what isotopes are??
Oops, changed it. Yes, I know what isotopes are. (different in number of neureons, but same number of protons)
I assert that with the data given,u cannot determine the number of protons in the nucleus...Therefore,i cannot see a way to solve the problem...
P.S.U have the three mass numbers,if u are allowed to use the table of elements,u can find the numbers you're interested in quite easily...Then why would they give you the info with Br2+??Since when Br forms compounds at valence 2?? :surprised
Dexter, that's [itex]Br_2^+ [/itex] ions that are being measured by the spectrometer. The way a mass spectrometer works is that it measures deflection by some known applied field.
Physicsss: Look up a Periodic table (no other way). The atomic number of Br is 35. Then you know how many protons there are in the molecule. From here, you can handle it. Assume that both atoms in one molecule are identical isotopes.
(I'm not certain if this is a valid assumption, but the lack of any odd mass numbers suggests it is okay.)
I'm still unsure on how to do this problem...can you give me some hints?
From the information given, you have the mass numbers ( total number of nucleons )
From the periodic table , you have the number of protons. So, since you know that all three have the same number of protons, you can determine the number of nutrons.
what would the number of neutrons they each have tell me?
Physicsss,what is the definition of mass number ?
You are finding the mass of 2 atoms in a diatomic Br molecule.
So what combination of N1 and N2 will give 158, 160 or 162.
Well the lowest 158 has to be 2*N1, then the heaviest mass 162 = 2*N2, and that leaves the other possibility that N1+N2=160.
Mass number is the number of protons plus the number of neutrons...
To Astronuc: I don't quite understand what you're trying to say...
Astronuc,there are 4 posibilities,not 3:
[tex] 79+79;79+81;81+81;80+80 [/tex]
No Dexter, there are only 3 : those which Astronuc suggested.
80-80 is not a possibility, because then one would also expect spectral lines for 159 and 161.
Who said there weren't spectral lines for 159 and 161???Did u read carefully the text of the problem...??
Now,if u haven't seen the computer generated graph/curve which would return zero for 159 & 161 (namely for isotope 80),then u cannot tell whether it exists or not.
In this case,it doesn't.The isotope 80 doesn't exist in nature ,as one could learn from
,but only judging the text of the problem & the key-word "peak",you (actunally i) could be mislead to erroneous conclusions...
Bad wording from the problem's inventor... :yuck:
And BTW,who said that radioisotopes could not form a covalent bond???Incidentally,the halflife for the "tricky no.80" is of the order of 17minutes,but other isotopes would not half themselves in days...
I'm afraid all this is only likely to confuse the OP more.
physicsss : Each molecule has 2 atoms. Each atom may be any one of the possible isotopes. For some molecule, the Mass of the molecule is M = M1 + M2, the sum of the atomic masses (or mass numbers) = P + N1 + P + N2 = 2P + N1 + N2. Similarly, for another molecules, M' = 2P + N1' + N2' and M" = 2P + N1" + N2", where P is the number of protons (fixed), and Ns are the number of neutrons in the two atoms (which may be the same or different).
Substituting the given numbers (and using the number of protons from the atomic number), you will get 3 equations. Write these down, and we'll show you what to do next.
2*35+2*N=158, where N=44
2*35+2*N=160, where N=45
2*35+2*N=162, where N=46
is that right?
Hmm....after some thinking...I came up with the following:
am i on the right track?
If you change your notation,then,yes..
how do you mean?
But all these 6 variables need not be different. In fact, if there are only two different number, you can see that you can make 3 different sums from them. Let the numbers be a, b. Then the different sums will be a+a, a+b and b+b, or 2a, a+b and 2b.
Try these in the above equations, and if you get the combination right, they will work.
Separate names with a comma.