Question

Iodine-131 is an isotope of iodine (Z = 53) used for the treatment of hyperthyroidism, as it is readily absorbed into the cells of the thyroid gland. With a half-life of 8 days, it decays into an excited Xenon-131 atom (Z = 54).

Part A What type of radioactive decay occurs in this process? Gamma decay Beta decay Alpha decay Electron capture SubmitMy AnswersGive Up Correct Part B The related isotope iodine-125 is used in brachytherapy. In what way does it differ from iodine-131? A different number of protons The number of neutrons The number of beta particles The number of electrons SubmitMy AnswersGive Up Correct Part C Which of the following is a true statement regarding the atomic mass during the decay from iodine-131 to xenon-131? Which of the following is a true statement regarding the atomic mass during the decay from iodine-131 to xenon-131? We expect the mass has decreased because xenon-131 has one more proton. We expect the mass has increased because xenon-131 has an extra proton. We expect the mass has decreased because xenon-131 has one less proton. We expect the mass has increased because xenon-131 has one less proton. SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part D What percentage of an iodine-131 sample remains after 30 days? What percentage of an iodine-131 sample remains after 30 days? 83 % 13 % 50% 7.4 % SubmitMy AnswersGive Up Part E The SI unit for activity (decays per second) is the Becquerel (Bq). A hospital receives a shipment of iodine-131 with an activity of 8 × 1010 Bq. After 16 days, there are several patients that need to be given doses of 1 × 109 Bq each. How many patients can be treated? The SI unit for activity (decays per second) is the Becquerel (). A hospital receives a shipment of iodine-131 with an activity of 8  10 . After 16 days, there are several patients that need to be given doses of 1  10  each. How many patients can be treated? 10 patients 40 patients 80 patients 20 patients SubmitMy AnswersGive Up

Answer 1

That's is a all part A,B,C,D,E completely Solution show on......

A) the equation is given by

131 I 53 ----> 131 Xe 54 + 0 e -1

so

this is a beta decay

B)

given iodine - 125

now consider iodine - 131

we can see that

mass number of both is different

now

mass number = number of neutrons + number of protons

as

number of protons is always constant for a given element

we can say that

number of neutrons is different for iodine -125 and iodine - 131

so

the answer is number of neutrons

C)

now consider the decay equation

131 I 53 ---> 131 Xe 54 + 0 e -1

now

number of protons = atomic number

number of neutrons = mass number - atomic number

so

for Iodine - 131

number of protons = atomic number = 53

number of neutrons = 131 - 53 = 78

for Xenon - 131

number of protons = atomic number = 54

number of nuetrons = 131 - 54 = 77

now

we know that

mass of nuclues = total mass of protons + total mass of nuetrons

so

mass of iodine nucleus = ( 53 x mass of proton ) + ( 78 x mass of nuetron)

mass of Xenon nucleus = ( 54 x mass of proton) + ( 77 x mass of nuetron)

now

consider

mass of iodine - mass of xenon = (53 x mass of proton)+(78 x mass of nuetron) - (54 x mass of proton)+(77 x mass of nuetron)

mass of iodine - mass of xenon = mass of neutron - mass of proton

now

we know that

mass of neutron > mass of proton

so

mass of neutron - mass of proton > 0

so

mass of iodine - mass of xenon = mass of neutron - mass of proton > 0

mass of iodine - mass of xenon > 0

mass of iodine > mass of xenon

now

131 I 53 ---> 131 Xe 54 + 0 e -1

as mass of iodine -131 > mass of Xenon- 131

there is a decrease in mass

so

the answer is

we expect the mass has decreased because xenon-131 has one more proton

D)

we know that

for radioactive decay

decay constant (k) = ln2 / half life

given

half life = 8 days

so

decay constant (k) = ln2 / 8

decay constant (k) = 0.0866434

now

for radioactive decay

N = No x e^(-kt)

N/No = e^(-kt)

given

time = 38 days

so

N/No = e^(-0.0866434 x 38)

N/No = e^(-3.292449)

N/No = 0.03716

(N/No) x 100 = 0.03716 x 100 = 3.716

so

3.7 % of the sample will remain after 38 days

E)

we know that

for radioactive decay

A = Ao x e^(-kt)

A/Ao = e^(-kt)

given

time = 16 days

so

A/Ao = e^(-0.0866434 x 16)

A/Ao = e^(-1.386294361)

A/Ao = 0.25

A = 0.25 Ao

given

intial activity (Ao) = 6 x 10^10

so

A = 0.25 x 6 x 10^10

A = 1.5 x 10^10

so

the activity that remains after 16 days is 1.5 x 10^10 Bq

now

number of patients that can treated = activity / dose

given

dose = 1 x 10^9

so

number of patients that can be treated = 1.5 x 10^10 / 1 x 10^9

number of petients that can be treated = 15

so

15 patients can be treated