Consider the hyperfine interaction of a hydrogen nucleus...

Consider the hyperfine interaction of a hydrogen nucleus with magnetic moment μ(N) = 2.79μ(K) of 33.5T. The separation of the hyperfine components is equal to: (a) 0.71GHz (b) 1.43GHz (c) 2.60GHz (d) 2.86GHz Consider the hyperfine interaction of a hydrogen nucleus with magnetic moment μ = 2.79k (where k is the nuclear magneton) with a bound electron having an internal magnetic field strength of 33.5 T. The separation of the hyperfine components is equal to: (a) 0.71GHz (b) 1.43GHz (c) 2.60GHz (d) 2.86GHz

Transcript

00:02 Okay, so in this problem we have an interaction between the magnetic depot moment of the proton and the spin magnetic depot moment of the electron.

00:16 There are splitting the ground level into two separated levels.

00:23 Each one of these levels with an energy, let's put here a delta e.

00:29 This is the splitting energy of 5 .9 times 10 to the minus 6 electron volts.

00:41 Okay, so we have this interaction splitting the ground level with this energy, and we must find what is the wavelength and the frequency of the emitted photon in this splitting interaction.

01:08 Okay, so let's begin with question a, with item a, which is to find out the wavelength and the frequency.

01:22 So, again, let's remember that the transition energy for a photon, that's called delta e, is described by the equation hc divided by lambda, where lambda is the wavelength of the photon.

01:41 Therefore, we are arranging this equation.

01:45 We have the wavelength that we want to find out is going to be equal, oops, hc divided by delta e.

01:57 Okay? well, this is basically 4 .136 times 10 to the minus 15.

02:10 This is electron volts seconds, okay, the planck constant, that multiplies the speed of light, which is 3 times 10 to the 8, divided by the energy, the splitting energy, which is 5 .9 times 10 to the minus 6.

02:35 Therefore, calculating this, we find out that the wavelength of the photon is of 21 centimeters...

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