Question

Can you help me to solve this problem ?​

Answer 1

1) 14 days :

Amount of P after 14 days = 45.98 mg

Amount of S = (125 -45.98) = 79.02 mg

2) 28 days

Amount of P = 16.916 mg

Amount of S = (125 - 16.916) = 108.08 mg

3) 42 day

Amount of P = 6.22 mg

Amount of S = (125 - 6.22) = 118.78 mg

4) 56 days

Amount of P-32 = 2.28 mg

Amount of S = (125 - 2.28) = 122.72 mg

Step-by-step explanation:

The formula used to calculate the amount of the radioactive substance after time t is :

`N = N_(0)e^(-\lambda t)`

Here ,

N = Amount of substance after time t

N0 = Initial amount of substance

t = time

`{\lambda}` = decay constant

`{\lambda} = (1)/(Half-Life)`

The equation for beta- decay is :

`_(15)^(32)\textrm{P}\rightarrow _(14)^(32)\textrm{S}+`
`_(1)^(0)\textrm{e}+\gamma`

Initially only P-32 is present = 124 mg

So

N0 = 125 mg

Half - life = 14 days

`{\lambda} = (1)/(14)`

`{\lambda} = 0.0714 day^(-1)`

a).

t = 14 days

`{\lambda t}` = 14 x 0.0714 = 1 day

`N = N_(0)e^(-\lambda t)`

`N = 125e^(-1)`

=45.98 mg

Amount of P after 14 days = 45.98 mg

This means from 125 mg we have 45.98 mg left . So (125 -45.98) has reacted to produce S

Amount of S = (125 -45.98) = 79.02 mg

b) 28 days

t = 28 days

`{\lambda t}` = 28 x 0.0714 = 2 day

`N = N_(0)e^(-\lambda t)`

`N = 125e^(-2)`

`e^(-2)` use scientific calculator to solve this

= 16.916 mg

Amount of P = 16.916 mg

Amount of S = (125 - 16.916) = 108.08 mg

c) 42 days

t = 42 days

`{\lambda t}` = 42 x 0.0714 = 3 day

`N = N_(0)e^(-\lambda t)`

`N = 125e^(-3)`

= 6.22 mg

Amount of P-32 = 6.22 mg

Amount of S = (125 - 6.22) = 118.78 mg

d) 56 days

t = 56 days

`{\lambda t}` = 56 x 0.0714 = 4 day

`N = N_(0)e^(-\lambda t)`

`N = 125e^(-4)`

= 2.28 mg

Amount of P-32 = 2.28 mg

Amount of S = (125 - 2.28) = 122.72 mg