A radioactive element has decayed to 1/32 of its original concentration in 30 days. What is the half-life of this element?

Question

A radioactive element has decayed to 1/32 of its original concentration in 30 days. What is the half-life of this element?

Answer 1

Let:

A = remaining radioactive element after some time, t

A_o = initial amount of the radioactive element at time = 0

t = time passed

t_1/2 = half-life, or the time required for half of the initial amount to decay, A = 1/2 * A_o

The exponential decay formula:

A = A_o * (1/2) ^{t / t1/2}

tells us how much of element A is leftover from an initial A_o amount after t time with respect to its half-life t_1/2

We are given:

A = 1/32 * A_o

t = 30 days

After 30 days, we have 1/32 of the original amount of element A we started with.

First, let's solve for the variable we need to calculate, t_1/2

A = A_o * (1/2) ^{t / t1/2}

Move the amount variables to the same side

A / A_o = (1/2) ^{t / t1/2}

Undo (1/2) to a power with a log base 1/2 on both sides

log_1/2 (A / A_o) = t / t_1/2

Isolate our variable of interest, t_1/2 and re-arrange

1 / t_1/2 = log_1/2 (A / A_o) / t

Simplify by taking the reciprocal of both side (1 / (1/x) = x for example)

t_{1/2 =} t / log_1/2 (A / A_o)

Now we are ready to plugin our knowns

t_1/2 = (30 days) / log_1/2 (1/32A_o / A_o)

A_o cancels out in the fraction inside the log

t_1/2 = 30 days / log_1/2 (1/32)

Now, we can calculate log_1/2 (1/32) in two ways:

1) First we can recognize that 2⁵ = 32 therefore 1/32 = (1/2)⁵.

This means:

log_1/2 (1/32) = log_1/2 (1/2)⁵ = 5 log_1/2 (1/2) = 5 * 1 = 5

and therefore:

t_1/2 = 30 days / 5 = 6 days

2) If we did not recognize that 1/32 is the 5th power of the base of our original log, 1/2, or we are in a situation where the remaining amount of the element is not an easy whole number power of the base, we can find the result of log_1/2 (1/32) using the change of base formula and use a calculator:

log_b (a) = log_x (a) / log_x (b)

In our case:

b = 1/2

a = 1/32

x = any log base that is convenient (most calculators have log base 10 or ln natural log (log base e).

So, using log₁₀ as log_x, we calculate log_1/2 (1/32) as follows:

log_1/2 (1/32) = log₁₀ (1/32) / log₁₀ (1/2) = 5

t_1/2 = 30 days / 5 = 6 days

Answer 2

The formula for the decay is:

R = M(0.5)^(t/h)

In this formula:

R is the remaining mass

M is the initial mass

t is the period of time elapsed

h is the half-life in the same units as t

The problem specifies that R=M/32 after 30 days, so we have:

M/32 = M(0.5)^(30/h)

1/32 = (0.5)^(30/h)

log_(1/2)(1/32) = 30/h

5 = 30/h

h = 30/5

h = 6

The half-life is 6 days.

Note that "log_(1/2)(1/32)" is the logarithm in base 1/2 of 1/32.

Answer 3

To find the half-life of the radioactive element, you can use the exponential decay formula:

\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T}} \]

where:

\( N(t) \) is the remaining quantity of the substance after time \( t \),

\( N_0 \) is the initial quantity of the substance,

\( T \) is the half-life,

\( t \) is the elapsed time.

Given that the substance decays to \(\frac{1}{32}\) of its original amount in 30 days, you can set up the equation:

\[ \frac{N_0}{32} = N_0 \left( \frac{1}{2} \right)^{\frac{30}{T}} \]

Dividing both sides by \( N_0 \):

\[ \frac{1}{32} = \left( \frac{1}{2} \right)^{\frac{30}{T}} \]

Recall that \( \frac{1}{32} = \left( \frac{1}{2} \right)^5 \). Therefore, you can equate the exponents:

\[ \left( \frac{1}{2} \right)^5 = \left( \frac{1}{2} \right)^{\frac{30}{T}} \]

This gives you:

\[ 5 = \frac{30}{T} \]

Solving for \( T \):

\[ T = \frac{30}{5} = 6 \]

So, the half-life of the radioactive element is 6 days.

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