Answer to Question #258523 in Chemistry for Leo

Question

The half-life of strontium-91, 91

38Sr is 9.70 h. Find

(a) its decay constant, and (5pts)

(b) for an initial 1.00-g sample, the activity after 15 hours.

Answer 1

Solution:

(a):

The time required for half of the original population of radioactive atoms to decay is called the half-life.

The relationship between the half-life (t_1/2) and the decay constant (λ) is given by t_1/2 = 0.693 / λ.

Hence,

λ = 0.693 / t_1/2 

λ = (0.693) / (9.70 h) = 0.0714 h^-1

λ = 0.0714 h^-1

(b):

Calculate the number of atoms in 1.00 g of strontium-91:

N_o = m × N_a / M

N_o = (1.00 g × 6.022×10²³) / (91 g/mol) = 6.618×10²¹

Now, the activity of the sample:

A = N_o × λ × e^–λt,

where:

A = ending activity

N_o = starting number of atoms

e = constant = 2.71828

λ = decay constant

t = elapsed time

Hence,

A = (6.618×10²¹) × (0.0714 h^-1) × (2.71828)^{–0.0714 × 15} = 1.62×10²⁰ Bq

A = 1.62×10²⁰ Bq

Answers:

(a): λ = 0.0714 h^-1

(b): A = 1.62×10²⁰ Bq