Question

Question 6. Please show your working for all questions. (a) If a radioactive nuclide decays for an interval of time equal to 4 times its half-life, what fraction of the original activity remains ? (b) A researcher is doing an experiment with a cell culture to image the DNA. To do so, the researcher applies ethidium bromide (EtBr) to the culture. After preparation, 50% of the EtBr has bound to DNA and 50% is in free solution. Imaging is done using fluorescence spectroscopy. The sample is illuminated by a short pulse of UV light that excites the EtBr molecules in the sample. DNA bound and free EtBr molecules have different fluorescence lifetimes. Unbound EtBr has a fluorescence lifetime of 1.7 ns, whereas DNA-bound EtBr has a lifetime of 23 ns. (i) How long will it take for the unbound EtBr fluorescence to decrease by 50%? (ii) What will be the fluorescence level (in %) of the DNA-bound EtBr after 5 ns? (iii) How long a delay is required until the background fluorescence (fluorescence from the free EtBr) is less than 1% of the total fluorescence intensity? (c) For a focal spot size of 2.5 mm and fixed source-object distance of 100 cm, what object-film distance would yield a geometric unsharpness of 0.45 mm? (d) A biopsy sample has been collected from a patient who is suspected to have skin cancer. What form of imaging would you use to determine if the cells collected are of healthy or cancerous tissue? Justify your answer. (e) You have been asked to determine the volume of the skull cavity and therefore the largest possible brain size of a newly discovered dinosaur. Which imaging technique would you recommend for this purpose? Justify your answer. (f) In biomedical imaging, we often need to keep the imaging time short. Briefly explain the trade- off that can be done to shorten the imaging time. Your answer should include which imaging technique you use as an example, why the imaging time must be kept short, and which other quality is therefore degraded and why.

Answer 1

(a) Decay constant = $\lambda$ = ( ln 2 ) / T, T being the half - life of the substance.

Hence, activity after time t = 4T is : A = $\lambda$ N_o exp ( - $\lambda$ t ), N_o being the initial amount of substance.

Original activity at t = 0 : A_o = $\lambda$ N_o.

Hence, fraction of activity remains after a time equal to 4 times of its half - life :

A / A_o = $\lambda$ N_o exp ( - $\lambda$ t ) / $\lambda$ N_o = exp ( - $\lambda$ t ) = exp [ { - ( ln 2 ) / T } x 4T ] = exp [ - ( ln 2 ) / 4 ] = 0.84.