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Chapter 35: Problem 2

Practise radioactivity interconversions. Express the following values in the alternative units indicated, with appropriate prefixes as necessary. Answer to three significant figures. (a) 72000 d.p.m. as Bq; (b) \(20 \mu \mathrm{Ci}\) as d.p.m.; (c) \(44400 \mathrm{~Bq}\) as \(\mu \mathrm{Ci}\); (d) \(6.3 \times 10^{5}\) d.p.m. \(\mathrm{mol}^{-1}\) as \(\mathrm{Bq} \mathrm{g}^{-1}\), for a compound with a relative molecular mass of 350 ; (e) 3108 d.p.m. as pmol, for a sample of a standard where the specific activity is stated as \(50 \mathrm{Ci} \mathrm{mol}^{-1}\).

Short Answer

Expert verified
The new activity values in the units indicated are as follows: (a) 1200 Bq (b) \(4.44 \times 10^{13}\) d.p.m. (c) \(1.20 \times 10^{-6} \, \mu\)Ci (d) 30 Bq g\(^{-1}\) (e) 0.01 pmol.

Step by step solution

01

Conversion of d.p.m. to Bq

To convert 72000 d.p.m. to Bq, use the relation 1 d.p.m. = 1/60 Bq: \((72000 \, \text{d.p.m.}) \times (1 \, \text{Bq} / 60 \, \text{d.p.m.}) = 1200 \, \text{Bq}\).
02

Conversion of \(\mu\)Ci to d.p.m.

To convert 20 \(\mu\)Ci to d.p.m., use the relations 1 \(\mu\)Ci = 3.7 x 10\(^{10}\) Bq and 1 Bq = 60 d.p.m.: \((20 \, \mu Ci) \times (3.7 \times 10^{10} \, Bq / \mu Ci) \times (60 \, \text{d.p.m.} / 1 \, \text{Bq}) = 4.44 \times 10^{13} \, \text{d.p.m.}\).
03

Conversion of Bq to \(\mu\)Ci.

To convert 44400 Bq to \(\mu\)Ci, use the relation 1 \(\mu\)Ci = 3.7 x 10\(^{10}\) Bq: \((44400 \, \text{Bq}) \times (1 \, \mu Ci / 3.7 \times 10^{10} \, \text{Bq}) = 1.20 \times 10^{-6} \, \mu Ci\).
04

Conversion of d.p.m. mol\(^{-1}\) to Bq g\(^{-1}\).

To convert \(6.3 \times 10^{5}\) d.p.m. mol\(^{-1}\) to Bq g\(^{-1}\) for a compound with a relative molecular mass of 350, first convert d.p.m. to Bq: \(6.3 \times 10^{5}\) d.p.m. \(\text{mol}^{-1}\) = \((6.3 \times 10^{5} / 60) \, \text{Bq mol}^{-1}\). Next, express the activity in Bq g\(^{-1}\) using the relative molecular mass: \((6.3 \times 10^{5} / 60) \, \text{Bq} \times \text{mol}^{-1} \times \text{mol} / 350 \, \text{g} = 30 \, \text{Bq g}^{-1}\).
05

Conversion of d.p.m. to pmol, using the given specific activity.

To convert 3108 d.p.m. to pmol, when the specific activity is given as \(50 \, \text{Ci} \times \text{mol}^{-1}\), first convert Ci to d.p.m. by using the conversion factors. 1 Ci = 3.7 x 10\(^{10}\) dps = 3.7 x 10\(^{10}\) x 60 d.p.m. Hence \(50 \, \text{Ci} \times \text{mol}^{-1} = 50 \, \text{mol}^{-1} Ci \times (3.7 \times 10^{10} \, \text{dps} / 1 \, \text{Ci}) \times (60 \, \text{d.p.m.} / 1 \, \text{dps}) = 3 \times 10^{13} \, \text{d.p.m.} \times \text{mol}^{-1}\). So, \(3108 \, \text{d.p.m.} \times \text{mol} / (3 \times 10^{13} \, \text{d.p.m.}) \times (\text{10}^{12} \, \text{pmol} / 1 \, \text{mol}) = 0.01 \, \text{pmol}\).

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Most popular questions from this chapter

Use the concept of specific activity in calculations. A researcher wishes to estimate the rate of uptake of the sugar galactose by carrot cells in a suspension culture. She prepares \(250 \mathrm{~mL}\) of the cell culture medium containing \(10^{7}\) cells per \(\mathrm{mL}\) and unlabelled galactose at a concentration of 5 mmol \(L^{-1}\). She then 'spikes' this with \(5 \mu \mathrm{L}\) (regard this as an insignificant volume) of radioactive standard containing \(55 \mathrm{MBq}\) of \({ }^{14} \mathrm{C}\)-labelled galactose (regard as an insignificant concentration). Answer to two significant figures. (a) Calculate the specific activity of the galactose in the culture solution in \(\mathrm{Bq} \mathrm{mol}^{-1}\). (b) If the total cell sample takes up \(79.2 \times 10^{5} \mathrm{~Bq}\) in a 2-h period, calculate the galactose uptake rate in \(\mathrm{mol} \mathrm{s}^{-1} \mathrm{cell}^{-1}\).
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