Question

Doubly‑charged calcium ions, Ca2+,Ca2+, are released into your bloodstream and flow through your capillaries. The diameter of a capillary is about 7.70 μm,7.70 μm, calcium ion concentrations are approximately 1.30 mol/m3,1.30 mol/m3, and typical electric current flow due to the Ca2+Ca2+ ions is around 1.2 nA.1.2 nA.

Calculate the drift speed ?driftvdrift of the calcium ions

Answer 1

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Notes vac = 102x10 solution | Given :- het, q= Change on each Calcium Ca ion ipatze . 9 = +2X16X 10 - 19 cb = 3=2 x 10 19 cb Is Electric current due to flow of Ca ions: n = Caklum con (Ca2+) concentration/ density in capillary = 1.30 mol/m² o n=103 x Avagadso noi /m3 n = 1-3x 6.023x10²3 / m² n=782 ggrola ions /m² :708299x10 m² liameter of capillary = 7.70 um=77x10m o cross section Area of Capillary is, A=Td A = 3.14% (7 78100 = 186•2658102 m² 4 drift speed of cart cons is given as to- Va = = 1.2 x 10-9 0 4 4 183 19 x 10^3^2 05 x 86- 165 X 1012 - 23 . V d = ♡.000 2 5 7x10 x 102 x 10the 10t12 = 0.000257 x 10 m/s = 25.71 lm, = 25.71 X 10 m/s Ang Aus