Question

Many radioisotopes have important industrial, medical, and research applications. One of these is 60Co, which has a half-life of 5.20 years and decays by the emission of a beta particle (energy 0.310 MeV) and two gamma photons (energies 1.17 MeV and 1.33 MeV). A scientist wishes to prepare a 60Co sealed source that will have an activity of at least 10.8 Ci after 42.0 months of use.

a) If the activity is 10.8 Ci, how many 60Co atoms are in the source? Tries 0/10

b) What is the minimum number of nuclei in the source at the time of creation? Tries 0/10

c) What is the minimum initial mass of 60Co required?

Answer 1

Answer-

Given,

half life = 5.20 years or 1.64 * 10⁸ sec   [1 year = 3.154 * 10⁷ seconds]

Activity = 10.9 Ci or 4.033 * 10¹¹ Bq [1 Ci = 3.7 * 10¹⁰ Bq]

Molar mass of 60Co = 59.933816 g/mol

time = 42 months or 1.104 * 10⁸ sec [1 Month = 2.63 * 10⁶ seconds]

a)

Number of atoms in 60 Co source = ?

We know that,

A = λ * N --------------------------3

where, A = activity in Bq

λ = decay constant

N = Number of Undecayed atoms in sample

Also,

half life = 0.693 / λ --------------4

From 3 and 4,

N = A / (0.693 /half life )

where, A = activity in Bq

N = Number of Undecayed atoms in sample

Put the values,

N = A / (0.693 /half life )

N = 4.033 * 10¹¹ Bq / (0.693 /1.64 * 10⁸ sec )

N = 9.54 * 10¹⁹ atoms

So, the number of atoms that will have activity of 10.8 Ci or 4.033 * 10¹¹ Bq is, 9.54 * 10¹⁹ atoms. these are the atoms left after 42 months i.e. 1.104 * 10⁸ sec [1 Month = 2.63 * 10⁶ seconds]

Also,

Mass = Number of atoms/avogadro's number) * molar mass

mass in 9.54 * 10¹⁹ atoms = (9.54 * 10¹⁹ atoms / 6.022 * 10²³) * 59.933816 g/mol

mass in 9.54 * 10¹⁹ atoms = 94.89 * 10^-4 g

Also,

= 0.693 / half life

where, = decay rate

decay rate of 60 Co = 0.693/1.64 * 10⁸

So, decay rate of 60 Co = 4.23 * 10^-9 Bq

Also,

N_t = Ne^-t

where, N_t = the amount of radioactive particles at time (t)

N₀ = the amount of radioactive particles at time = 0

λ = rate of decay constant

t = time

Put the values,

N_t = Ne^-t

94.89 * 10^-4 g = (N) * e^{-0.00000000423 * 110400000}

94.89 * 10^-4 g = (N) * 0.627 [e^-0.467 = 0.627]

N = 151.34 * 10^-4 g

Number of atoms in 151.34 * 10^-4 g of 60 Co is,

Number of atoms = (Mass/molar mass) * avogadro's number

Number of atoms = (151.34 * 10^-4 g / 59.933816 g/mol) * 6.022 * 10²³

Number of atoms = 1.521 * 10²⁰ atoms

So, there are 1.521 * 10²⁰ atoms present in the source due to which the activity of remaining atoms is 10.8 Ci after 42 months.

b)

Minimum Number of nuclie in the source = ?

The atoms needed to have Activity of 10.8 Ci after 42 months of use is 1.521 * 10²⁰ atoms. So, the minimum number of atom or nuclei in source at the time of creation is same 1.521 * 10²⁰ atoms.

c)

Number of atoms in the souce = 1.521 * 10²⁰ atoms

Mass of atoms in the source = ?

We know that,

Mass = Number of atoms/avogadro's number) * molar mass

Put the values,

Mass = (1.521 * 10²⁰ atoms / 6.022 * 10²³ ) * 59.933816 g/mol

Mass = 1.514 * 10^-2 g

So, the minimum mass in the source is 1.514 * 10^-2 g