A sample contains 1010 atoms of a radioactive isotope with a half-life of 30 days. Calculate (i) the initial activity in disintegrations/sec and (ii) the time for the number of atoms to reduce to 104.

Question

Asked on April 27, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-27T08:46:32.629Z

Step 1: Identify the given values and convert units. The initial number of atoms is N_0 = 10^10 atoms. The half-life is T_1/2 = 30 days. To work with seconds for activity calculations, convert the half-life to seconds: T_1/2 = 30 days × 24 hoursday × 60 minuteshour × 60 secondsminute T_1/2 = 2,592,000 s Step 2: Calculate the decay constant . The decay constant is related to the half-life by the formula: = ((2))/(T_1/2) = (0.693147)/(2,592,000 s) ≈ 2.67495 × 10^-7 s^-1 i) Calculate the number of disintegrations in the first second. The number of disintegrations per second is the activity A. At the initial instant, this is A_0. A_0 = N_0 A_0 = (2.67495 × 10^-7 s^-1) × (10^10 atoms) A_0 = 2674.95 disintegrations/second Rounding to four significant figures: 2.675 × 10^3 disintegrations/second ii) What time will elapse before 10^4 atoms remain? We use the radioactive decay formula: N(t) = N_0 ((1)/(2))^t/T_1/2 Given N(t) = 10^4 atoms, N_0 = 10^10 atoms, and T_1/2 = 30 days. 10^4 = 10^10 ((1)/(2))^t/30 Divide both sides by 10^10: (10^4)/(10^10) = ((1)/(2))^t/30 10^-6 = ((1)/(2))^t/30 Take the natural logarithm of both sides: (10^-6) = (((1)/(2))^t/30) -6 (10) = (t)/(30) ((1)/(2)) Since ((1)/(2)) = -(2): -6 (10) = -(t)/(30) (2) 6 (10) = (t)/(30) (2) Solve for t: t = 30 days × (6 (10))/((2)) t = 30 days × (6 × 2.302585)/(0.693147) t = 30 days × (13.81551)/(0.693147) t = 30 days × 19.93157 t = 597.947 days Rounding to one decimal place: 597.9 days iii) What is the Count rate at this time? "This time" refers to when N(t) = 10^4 atoms. The count rate is the activity A(t) at this specific time. A(t) = N(t) Using the decay constant ≈ 2.67495 × 10^-7 s^-1 and N(t) = 10^4 atoms: A(t) = (2.67495 × 10^-7 s^-1) × (10^4 atoms) A(t) = 2.67495 × 10^-3 disintegrations/second Rounding to four significant figures: 2.675 × 10^-3 disintegrations/second Send me the next one 📸

Accepted Answer · 2026-04-27T08:46:32.629Z

Step 1: Identify the given values and convert units. The initial number of atoms is N_0 = 10^10 atoms. The half-life is T_1/2 = 30 days. To work with seconds for activity calculations, convert the half-life to seconds: T_1/2 = 30 days × 24 hoursday × 60 minuteshour × 60 secondsminute T_1/2 = 2,592,000 s Step 2: Calculate the decay constant . The decay constant is related to the half-life by the formula: = ((2))/(T_1/2) = (0.693147)/(2,592,000 s) ≈ 2.67495 × 10^-7 s^-1 i) Calculate the number of disintegrations in the first second. The number of disintegrations per second is the activity A. At the initial instant, this is A_0. A_0 = N_0 A_0 = (2.67495 × 10^-7 s^-1) × (10^10 atoms) A_0 = 2674.95 disintegrations/second Rounding to four significant figures: 2.675 × 10^3 disintegrations/second ii) What time will elapse before 10^4 atoms remain? We use the radioactive decay formula: N(t) = N_0 ((1)/(2))^t/T_1/2 Given N(t) = 10^4 atoms, N_0 = 10^10 atoms, and T_1/2 = 30 days. 10^4 = 10^10 ((1)/(2))^t/30 Divide both sides by 10^10: (10^4)/(10^10) = ((1)/(2))^t/30 10^-6 = ((1)/(2))^t/30 Take the natural logarithm of both sides: (10^-6) = (((1)/(2))^t/30) -6 (10) = (t)/(30) ((1)/(2)) Since ((1)/(2)) = -(2): -6 (10) = -(t)/(30) (2) 6 (10) = (t)/(30) (2) Solve for t: t = 30 days × (6 (10))/((2)) t = 30 days × (6 × 2.302585)/(0.693147) t = 30 days × (13.81551)/(0.693147) t = 30 days × 19.93157 t = 597.947 days Rounding to one decimal place: 597.9 days iii) What is the Count rate at this time? "This time" refers to when N(t) = 10^4 atoms. The count rate is the activity A(t) at this specific time. A(t) = N(t) Using the decay constant ≈ 2.67495 × 10^-7 s^-1 and N(t) = 10^4 atoms: A(t) = (2.67495 × 10^-7 s^-1) × (10^4 atoms) A(t) = 2.67495 × 10^-3 disintegrations/second Rounding to four significant figures: 2.675 × 10^-3 disintegrations/second Send me the next one 📸

A sample contains 1010 atoms of a radioactive isotope with a half-life of 30 days. Calculate (i) the initial activity in disintegrations/sec and (ii) the time for the number of atoms to reduce to 104.

Need help with your own homework?

More Physics Questions

A sample contains 1010 atoms of a radioactive isotope with a half-life of 30 days. Calculate (i) the initial activity in disintegrations/sec and (ii) the time for the number of atoms to reduce to 104.

Need help with your own homework?

More Physics Questions