Question

Bismuth-210 is beta emitter with a half-life of 5.0 days. Part A If a sample contains 1.2 g of Bi−210 (atomic mass = 209.984105 amu), how many beta emissions would occur in 14.0 days? Part B If a person's body intercepts 5.8% of those emissions, to what dose of radiation (in Ci) is the person exposed?

Answer 1

we know that

moles = mass / molar mass

so

moles of Bi = 1.2 / 209.984105

moles of Bi = 5.714718264 x 10-3

now

we know that

number of atoms = moles x avagadro number

so

number of Bi atoms (No) = 5.714718264 x 10-3 x 6.023 x 10^23

number of Bi atoms (No) = 3.4419748 x 10^21

now

decay constant ( lamda ) = ln2 / half life

so

lamda = ln2 / 5

now

ln N = ln No - ( lamda x t)

so

ln ( N/No) = - ( ln 2 ) x 14 /5

N/No = 0.143587

N = 0.143587 x 3.44 x 10^21

N = 4.942238488 x 10^20

now

number of emission = No - N

= 3.4419748 x 10^21 - ( 4.942238488 x 10^20)

= 2.947750951 x 10^21

so

2.94775 x 10^21   beta emissions occur in 14 days

B)

now

5.8 % of those emissions are

emissions intercepted = 0.058 x 2.94775 x 10^21

emissions intercepted = 0.170969555 x 10^21

now

time = 14 days = 14 x 24 x 60 x 60 = 1209600

now

emirrion intercepted / time = 0.17096955 x 10^21 / 1209600

= 1.413438781 x 10^14

we know that

1 Ci = 3.7 x 10^10 emission per second

so

dose = 1.413438781 x 10^14 / 3.7 x 10^10

dose = 3820 Ci

so

the dose of radiation is 3820 Ci