 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/2012/s16/1/2/solution

Define the one-dimensional thermal widths

σ_{p} = k_{B} T_{p} / m_{e}

and

σ_{b} = k_{B} T_{b} / m_{e}

f_{p} (v) = \frac{n _{p}}{2 π σ _{p}} e^{- v^{2} / (2 σ_{p}^{2})}, f_{b} (v) = \frac{n _{b}}{2 π σ _{b}} e^{- (v - V)^{2} / (2 σ_{b}^{2})} .

The total is

f = f_{p} + f_{b}

. It consists of a broad peak centred at zero and a much narrower displaced beam near

V

, with

V ≫ σ_{p} ≫ σ_{b}

. Their areas, rather than their peak heights, are

n_{p}

and

n_{b}

. A weak narrow beam can make a positive-slope region on its low-velocity side even while contributing little to the total density.

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