 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015/s10/3/1/solution

T = 300 K

Δ E = (1.054 \times 1 0^{- 34}) (2.675 \times 1 0^{8}) (1.5) = 4.23 \times 1 0^{- 26} J,

\frac{Δ E}{k _{B} T} = \frac{4.23 \times 1 0 ^{- 26}}{( 1.380 \times 1 0 ^{- 23} ) ( 300 )} = 1.02 \times 1 0^{- 5} .

293 K

the ratio is

1.05 \times 1 0^{- 5}

; the room-temperature convention makes only a small difference. The corresponding fractional nuclear spin polarization has magnitude about five parts per million. The source's negative sign convention assigns the label up to the higher-energy state; it should not be confused with the lower-energy magnetic moment aligned with a positive proton magnetic field.

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