Question

A sample of radium has a weight of 100 mg and a half-life of

Answer 1

Answer:

1a ) N_6 = 50 mg

1b) N_3 = 70.7 mg

1c) N_1 = 89.1 mg

2) f(t) = 100e^(-0.1155t)

Step-by-step explanation:

Formula to get amount remaining is;

N_t = N_o × ½^(t/t_½)

We are given;

N_o = 100 mg

t_½ = 6 years

We have; N_t = 100(½^(t/6))

Thus;

1a) amount left after 6 years is;

N_6 = 100 × ½^(6/6)

N_6 = 50 mg

1b) amount left after 3 years is;

N_3 = 100 × ½^(3/6)

N_3 = 70.7 mg

1c) amount left after 1 year is;

N_1 = 100 × ½^(1/6)

N_1 = 89.1 mg

2) to find the function to model the amount left after t years, let's use the formula;

f(t) = f_o × e^(-rt)

Where r is decay constant

Thus;

f(t) = 100e^(-rt)

We recall that N_t = 100(½^(t/6))

Thus;

100(½^(t/6)) = 100e^(-rt)

Divide both sides by 100 to get;

(½^(t/6)) = e^(-rt)

In (½^(t/6)) = In e^(-rt)

Using power rule, we have;

(t/6) In ½ = -rt In e

-0.6931t/6 = - rt (because In e = 1)

t will cancel out to give;

r = 0.6931/6

r = 0.1155

Thus;

f(t) = 100e^(-0.1155t)