A form of MRI (usually called nuclear magnetic resonance) is used a great deal t

Question

A form of MRI (usually called nuclear magnetic resonance) is used a great deal to analyze what atoms are in an unknown substance. Every chemistry department has several NMR machines for carrying out chemical analysis. The nuclear magnetic moment of a nitrogen atomic nucleus is about 1/14 th as large as the spin of a hydrogen nucleus, and for a sodium nucleus the magnetic moment is about 1/4 as big as for hydrogen. How could you modify an MRI machine to make it into an NMR machine that would detect the amount of nitrogen and sodium atoms in a chunk of material relative to the amount of hydrogen To modify the MRI to measure amount of Nitrogen, which of the following changes would work (we are making only one change at a time)? (check all that apply) To modify the MRI to measure the amount of Sodium, which of the following changes would work (we are making only one change at a time)? (check all that apply) If radio waves of 1.5 MHz are absorbed by unknown atoms in a gas, what is the energy difference (in eV) between the spin-up and spin-down states? If this is the energy required to flip the spin in an applied magnetic field of 0.035T, what is the magnetic moment? What atoms are in the gas? (enter periodic table symbol)

Explanation / Answer

ANSWER:

2.

a. The energy of nucleus is directly proportional nulcler magnetic moment (µ) and external magnetic filed strength (B)

E = - µB

since µ of N is 1 / 14th of Hydrogen therefore magnetic filed strength (B0 must INCREASED 14 times

Furher the frequency required to excite the nucleus is also proportional to B. Hence RADIOWAVES WITH 14 TIMES the energy must be used

Hence 2nd and 6th change will work

b. Same applies to sodium. 4th and last change will work.

c. Energy difference = hv

h = planks constant = 6.62 X 10-34 kg m2s-1

v = frequency = 1.5MHz = 1.5 X 106 Hz

Energy difference = 6.62 X 10-34 X 1.5 X 106 = 9.93 X 10-28 J = 9.93 X 10-28 X 6.24X 1018 = 6.2 X 10-9eV

1J =6.24X 1018eV

i) E = - µB

µ = - E / B = 9.93 X 10-28 / 0.035 = 2.83 X 10-26 J/T

ii) H2, He, N2 ,O2, F2, Cl2, Ne , Ar, Kr , Xe, Rn

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