Question

Refer the attached picture.

Proper solution required. [ Easy, but explanation needed.] ​

Answer 1

Answer :

4 days (Hope I am right ._., bit rusty with these stuff)

Explanation:

First we need to find the half life of the sample, which is given by

T1/2 = 0.693/λ

therefore, λ = 0.693/0.3465

λ = 2 days (since the unit of λis given by per day)

Now what we need to find is the no. of half lives required for the sample to become 1/4th of the initial sample.

In this case logically we can directly take it as 2 half lives as it is 1/4th of the initial... U can try calculating it for much lager no. by taking

$\frac{1}{4} = \frac{1}{2}^{n}$

which will give u the answer as 2.

Now we can multiply the no. of half lives required by the length of a single half life (2 days) which leaves you with the answer 4 days.

Please forgive me if there are any mistakes. Also make sure to ask any unclear parts.

Answer 2

Answer:

4 days

Explanation:

initial amt: N0 : 100

final amt : N: 100-75 = 25

k = 0.3465/day

N = N0 [e^(-kt)]

25 = 100 [e^(-0.3465t)]

log (100/25) = log [e^(0.3465t)]

log 4 = 0.3465 t

t = 1.3862 /0.3465

t = 4 days