Using the table of magnetic properties of the nuclei from the lecture notes, cal
ID: 474018 • Letter: U
Question
Using the table of magnetic properties of the nuclei from the lecture notes, calculate base resonance frequencies for 31P in two magnets, one producing magnetic field of 14.1 T, and the other operating at 900 MHz proton (1H) resonance frequency
Properties of selected nucleia Y (T s) Natural abundance Nucleus 2.6752 x 10 99.99 1 /2 4.107 x 107 0.012 13 6.728 x 10 1.07 1/2 14 1.934 x 10 99.63 2.713 x 10 15 0.37 1/2 3.628 x 10 17 5/2 0.038 2.518 x 10 19 100.00 1/2 7.081 x 10 23 100.00 3/2 Na 1.0839 x 10 31 100.00 1/2 l 13 Cd 5.961 x 10 12.22 1/2 aShown are the nuclear spin angular momentum quantum number, I, the magnetogyric ratio, y, and the natural isotopic abundance for nuclei of particular importance in biological NMR spectroscopy.Explanation / Answer
Dear Student
We know
n = (g/2p)B
where n = frequency , g = gyromagnetic ratio p= pi and B = magnetic field
It comes as energy = Planks constant * frequency and again energy = (hg/2p)B
In the given problem for 31P gyromagnetic ratio = 1.0839*108 magnetic field = 14.1 T, the value of pi = 3.14
So frequency = (1.0839*108 *14.1)/2*3.14 = 2.433*108 Hz = 243.3 MHz
The base frequency of the magnet is 243.3 MHz
If the operating frequency = 900 MHz
Then
900*1000000 = (1.0839*108 *B)/2*3.14
So B = 900*1000000*2*3.14/1.0839*100000000 = 52.14 T
So answer = 52.14 T
As frequency is directly proportional with the magnetic field so when magnetic field = 14.1 then frequencyis nearly 245Hz. When frequency is 900 i.e. 4 times than that of the previous frequency so magnetic field will also be four times and is is nearly 52 T.
Frequency = 243.3 Hz magnetic field = 14.1 T and
frequency = 900 Hz magnetic field = 52.14 T
Regards
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