Question

Can you please ensure that part e) is completed!

An isotope sometimes used in medicine is iodine 123, which has a half-life T1/2=13.22 hours.

a) What is the decay constant?

b) If the sample has an activity of 3×104 Bq, how many atoms are in the sample?

c) What is the activity of the sample in (b) 24 hours later?

d) The decay process occurs by electron capture. Give the reaction showing the decay products.

e) Iodine has a strong UV emission line at 183.0 nm when excited in an electrical discharge. What is the energy of associated atomic transition, expressed in eV.

Answer 1

(a) Decay constant = = ln2 / T1/2 = ln2 / ( 13.22 24 3600 ) = 6.068 10-7 s-1

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(b) dN/dt = N

where (dN/dt) activity in Bq ( Becquerel ) and N is umber of atoms present

N = (dN/dt) / = ( 3 104 ) / ( 6.068 10-7 ) = 4.944 1010

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(c) activity after 24 hours decreases by a factor

Hence activity after 24 hours = 3 104 0.284 = 8524 Bq

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(d) Iodine-123 decay scheme

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(e) Photon energy of UV emission at 183 nm, E = ( h c ) /

E = ( 6.626 10-34 3 108 ) / ( 183 10-9 ) = 1.086 10-18 J

Energy in eV , E = 1.086 10-18 / ( 1.602 10-19 ) = 6.78 eV