In an NMR experiment, the magnetic moment of a hydrogen nucleus (proton) points

Question

In an NMR experiment, the magnetic moment of a hydrogen nucleus (proton) points along the direction of an applied, external magnetic field of 0.5 T. A photon with a frequency of 2.13 times 10^7 Hz (near AM radio frequencies) is absorbed by the hydrogen nucleus flipping the direction of its magnetic dipole by 180 degree. Hence, the magnetic dipole moment of a proton is _____ A middot m^2. [The energy of a photon with a frequency f is E = hf. where h = 6.63 times 10^-34 J middot s (Plancks constant).] (a) 7.1 times 10^-27 (b) 1.4 times 10^-26 (c) 2.8 times 10^-26 (d) 5.6 times 10^-26 (e) 7.1 times 10^-26

Explanation / Answer

in the above problem, given terms are

Magentic field (B) = 0.5 T

Frequency of photon (f) = 2.13 x 10^7 hz

planks constant (h) = 6.63 x 10^-34

we usually know the formula of magnetic dipole is u= i*A

as it is given in the question that the magetic dipole is flipped by 180 degrees. An extra energy is required to keep the dipole moment in that direction.

extra energy (E )= h*f

E is realted with magnetic moment and magnetic field as E = - u.B ( vector dot product )

here '-' sign indiacte that magnetic moment is in opposite direction to B

h*f = - uBcos180

6.63 x 2.13 x 10^-34 x 10^7 = - (u x 0.5 x -1) {cos 180 = -1]

u = 14.1219 x 10^-27 / 0.5

u = 28.56 x 10*-27

magneticdipole moment of proton (u) = 2.8 x 10*-26 Am^-2

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