Question

A certain element undergoes beta-plus decay (also sometimes known as

Answer 1

The radioactive decay equation:

Activity is given by:

A_0 is the initial activity at time t = 0s

A(t) is the activity at time t

It can also be written in terms of No. of nuclei present at time t:

let t₁ = March 1. 2000

t₂ = April 2000

t₃ = March 2002

Now, given that:

at t₁ : No. of decays per hour = 183500 /hr = A(t₁)

at t₂ : No. of decays per hour = 181300 /hr = A(t₂)

We assume t1 as the starting point and then t = 31 days, A_0 = A(t_1), A(t) = A(t_2)

the relation becomes:

or,

Taking log both sides:

or,

or half life = 1775.45/(365) = 4.86 years

Now, We need to find N(t₁): No. of undecayed nuclei at t₁

The relation between decay rate and no. of undecayed nuclei :

or,

Substitute in above equation: (Convert activity rate to per second from per hour, and half life to seconds)

A(t) = 183500 /hr = 183500/3600 /s

t_1/2 = 1775.45 * 24 * 3600 s

2. how many decays per hour would you expect to measure on March 1, 2002?

Assuming we start from 1 March 2000, 1 March 2002 is t=2 yr from initial time.

So activity will be :