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A narrow stream of vanadium atoms in the ground state .^(4)F_(3//2) is passed through a transverse strongly inhomogeneous magnetic field of lengthl_(1)=5.0 cm as in the Stern-Gerlach experiment. The beam splitting is observed on a screen located at a distance l_(2)=15 cm from the magnet. The kinetic energy of the atoms is T=22 meV. At what value of the gradient of the magnetic field induction B is the distance between the extreme components of the split beam in the screen equal to delta=2.0 mm ? |
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Answer» Solution :In the homogeneous magnetic field the atom experience as force `F=gJmu_(B)(delB)/(delƵ)` Depending on the sign of `J`, this can be EITHER UPWARD or downward. Suppose the latter is true. The atom then traverses first along a parabola inside the field and , once outside, in a straight line. The toal distance between extreme lines on hte screen will be `delta=2gJmu_(B)(delB)/(delƵ){(1)/(2)((l_(1))/(V))^(2)+(l_(1))/(V).(l_(1))/(V)}//m_(V)` Here `m_(V)` is the MASS of hte vanadium atom. (The first term is the displacement within the field and the second term is the displacement due to the transverse velocity acquired in the magnetic field). Thus using `(1)/(2)m_(v)V^(2)=T`  we get `(delB)/(delƵ)=(2Tdel)/(gmu_(B)Jl_(1)(1_(1)+2l_(2)))` For vanadim atom in the ground state `.^(4)F_(3//2)`. `g=1+((3xx4)/(4)+(3xx5)/(4)-3xx4)/(2xx(3xx5)/(4))=1+(30-48)/(30)=1-(18)/(30)=(2)/(5)` `J=(3)/(2)`, using other data, and substituting we get `(delB)/(delƵ)=1.45xx10^(13)G//m` This VALUE differs from the answer given in the BOOK by almost a factor of `10^(9)`. For ncutral atoms in sterm Gerlach experimeters, the value `T=22 meV` is much too large. A more appropriate value will be `T=22 meV` i.e., `10^(9)` times smaller. Then one gets the right answer. |
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