Question

Suppose a radioactive substance decays at a rate of 4% per hour. How long does it take until the substance is half of its original weight?
7 days and 5 hours
1713 hours
712 hours
17 days and 8 hours

also have a few questionsmore.

Answer 1

Answer:

0.22 hours

Step-by-step explanation:

Half life is defined as the time taken by a radioactive material to decay to half of its original value.

Mathematically, half-life (t1/2) = ln2/¶ where ¶ is the decay constant.

Before we can get the half life, we need to get the decay constant first. To get ¶, we will use the radioactivity formula. According to the formula,

N/No = e^-¶t where;

N/No is the fraction of the final substance after decay the the original value of the substance = 4% = 1/25

t is the time of decay = 1hour

Substituting into the formula to get the decay constant we have;

1/25 = e^-1¶

Applying ln to both sides, we have;

ln 1/25 = lne^-¶

ln1/25 = -¶

-3.22= -¶

¶ = 3.22

The decay constant is 3.22

Half life = ln2/3.22

Half life = 0.22hours

It will take the substance 0.22hours before it changes to half of its original weight.

Answer 2

Answer:

The radioactive substance will decay in 17.70 hours .

Step-by-step explanation:

Given as :

The rate of decay of radioactive substance = 4 %  per hour

Let The original weight of substance = w

The final weight of substance = $\frac{w}{2}$

Let the hours for decay of substance = n

Now,

The weight of substance after n years = The original weight of substance × $(1- \dfrac{\textrm Rate}{100})^{n}$

or,

$\frac{w}{2}$ = w × $(1- \dfrac{\textrm 4}{100})^{n}$

Or, $\frac{1}{2}$ = $(1- \dfrac{\textrm 1}{25})^{n}$

or,  $\frac{1}{2}$ = $(\dfrac{\textrm 25-1}{25})^{n}$

or,  $\frac{1}{2}$ = $(\dfrac{\textrm 24}{25})^{n}$

Taking log both sides :

Log  $\frac{1}{2}$ = Log $(\dfrac{\textrm 24}{25})^{n}$

So, Log $\frac{1}{2}$ = n × Log $\frac{24}{25}$

∴ - 0.301 =  n × ( - 0.017 )

So , n = $\frac{0.301}{0.017}$

I.e n = 17.70  hours

Hence The radioactive substance will decay in 17.70 hours . Answer