While working in a magnetics lab, you conduct an experiment...

While working in a magnetics lab, you conduct an experiment in which a hydrogen atom in the $n$ = 1 state is in a magnetic field of magnitude $B$. A photon of wavelength $\lambda$ (in air) is absorbed in a transition from the $m$$_s$ = -$\frac{1}{2}$ to the $m$$_s$ = + $\frac{1}{2}$ state. The wavelengths $\lambda$ as a function of $B$ are given in the table. (a) Graph the data in the table as photon frequency $f$ versus $B$, where $f$ = c/$\lambda$. Find the slope of the straight line that gives the best fit to the data. (b) Use your results of part (a) to calculate $\mid$$\mu$$_z$$\mid$, the magnitude of the spin magnetic moment. (c) Let $\gamma$ = $\mid$$\mu$$_z$$\mid$$/$$\mid$$S$$_z$$\mid$ denote the gyromagnetic ratio for electron spin. Use your result of part (b) to calculate $\gamma$. What is the value of $\gamma$$/$($e$$/$2$m$) given by your experimental data?

While working in a magnetics lab, you conduct an experiment in which a hydrogen atom in the $n$ = 1 state is in a magnetic field of magnitude $B$. A photon of wavelength $\lambda$ (in air) is absorbed in a transition from the $m$$_s$ = -$\frac{1}{2}$ to the $m$$_s$ = + $\frac{1}{2}$ state. The wavelengths $\lambda$ as a function of $B$ are given in the table.
(a) Graph the data in the table as photon frequency $f$ versus $B$, where $f$ = c/$\lambda$. Find the slope of the straight line that gives the best fit to the data. (b) Use your results of part (a) to calculate $\mid$$\mu$$_z$$\mid$, the magnitude of the spin magnetic moment. (c) Let $\gamma$ = $\mid$$\mu$$_z$$\mid$$/$$\mid$$S$$_z$$\mid$ denote the gyromagnetic ratio for electron spin. Use your result of part (b) to calculate $\gamma$. What is the value of $\gamma$$/$($e$$/$2$m$) given by your experimental data?

Transcript

00:02 So for part a, it wants us to graph this, but we need to first find the frequency for each wavelength in the table with the problem.

00:13 So for the first column, that c divided by wavelength, which would be 2 .998 times 10 to the 8 meters per second, divided by 0 .024 meters.

00:31 Meters and that gives you 1 .40 times 10 to the 10th hertz so doing this for all the wavelengths gives the values in this table so i'm just going to write out the rest of the values for the table so we have b of t and then we have frequency of 10 to the 10th hertz so 0 .51 which is 1 .401 0 .74, which is 2 .097, 1 .03, which is 2 .802, 1 .52, which is 4 .199, 2 .02, 5 .604, 2 .08, be 7 .00, 5 .604, 2 .48, be 7 .00005, and then 2 .977 .97.

01:52 8 .398.

01:56 So we need to put these numbers on a graph for us.

02:02 So the graph will be like this.

02:10 Over here will be the frequency.

02:14 So 10 to the 10 hertz.

02:17 Down here will be b of t.

02:21 So it starts with zero.

02:23 Then it goes one, two, three, four, five, six, seven, eight, nine.

02:30 And 10.

02:34 Actually, it stops at 9, but okay, and then down here it is 0 .0 .5, 1, 1 .5, 2, 2 .5, and 3.

02:50 So if you, let's put a little mark here.

02:57 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7, 7, and nine so at 0 .5 you go up a little bit above one you're right there at in between these two 0 .5 and 1 you go up to 2 a little bit past 1 you go right below 3 a little bit past 1 .5 you go above 4 a little bit past 1 .5 you go above 4 a little bit past two, you're in between 5 and 6...

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