Question

Can you help me to solve this problem ?​

Answer 1

Answer:

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

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

${\lambda} = \frac{1}{Half-Life}$

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

${\lambda} = \frac{1}{14}$

${\lambda} = 0.0714 day^{-1}$

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