Barium-141 has a half-life of 18 minutes and a decay constant of 6.4 × 10-4 s-1.

Question 9

One possible nuclear fission reaction is

Barium-141 (¹⁴¹₅₆Ba) and krypton-92 (⁹²₃₆ Kr) are both β-emitters.

Barium-141 has a half-life of 18 minutes and a decay constant of 6.4 × 10^-4 s^-1.

The half-life of krypton-92 is 3.0 seconds.

(a) State what is meant by decay constant. [2]

(b) A mass of 1.2 g of uranium-235 undergoes this nuclear reaction in a very short time

(a few nanoseconds).

(i) Calculate the number of barium-141 nuclei that are present immediately after the

reaction has been completed. [2]

(ii) Using your answer in (b)(i), calculate the total activity of the barium-141 and the

krypton-92 a time of 1.0 hours after the fission reaction has taken place. [4]

Reference: Past Exam Paper – November 2013 Paper 43 Q8

Solution:

(a)

The decay constant is the probability of decay of a nucleus per unit time.

(b)

(i)

{One uranium nucleus decays into one barium nucleus (and other products).

One mole of uranium-235 contains 235 g of it.

And one mole of an element contains 6.02×10²³ particles (nucleus) of the element.

235 g - - - > 6.02 × 10²³ nuclei

12 g - - - > (1.2 × 6.02 × 10²³) / 235 nuclei of uranium

Each of these nuclei decay into barium nuclei.}

number = (1.2 × 6.02 × 10²³) / 235   

number = 3.1 × 10²¹                          

(ii)

{The number of nuclei present at time t may be obtained by:}

N = N₀ e^–λt

{Since the half-life of krypton is only 3 s, it would have negligible activity after 1 hour.}

- negligible activity from the krypton

{For barium,

N = N₀ e^–λt

As the uranium decays, it produces barium nuclei. So, the initial number of barium nuclei is the same as the number of uranium nuclei.

N₀ = 3.1 × 10²¹

Time t = 1 hour = 3600 s

λ = 6.4×10^-4 s^-1}

{With time, barium nuclei also decay. After time t, the remaining number of barium nuclei is give by:}

- for barium, N = (3.1 × 10²¹) exp(–6.4×10^–4 × 3600)

N = 3.1 × 10²⁰            

activity A = λN

A = 6.4 × 10^–4 × 3.1 × 10²⁰    

A = 2.0 × 10¹⁷ Bq