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Chapter 5: Q.5.76 (page 165)

If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 0.4 mg to 0.1 mg in 26.4 h, what is the half-life of iodine-123?

Short Answer

Expert verified

The half-life of iodine- $123$ is $13.2 h$

Step by step solution

01

Given Information

The amount of radioactive iodine- $123$ , used to treat thyroid cancer, in a sample decreases from $0.4$ mg to $0.1$ mg in $26.4$ h.

02

Explanation

We know,

The initial concentration of I- $123$ $N_{o}$ = $0.4$ mg and the concentration after $26.4$ h N = $0.1$ mg

Time t = $26.4$ h

Using, $In N - In N_{0} = - λ t$

$In \frac{0.1}{0.4} = - λ \times 26.4 h$

role="math" localid="1653899395922" $\Rightarrow λ = 0.05265 / h$

The half-life is

role="math" localid="1653899475118" $t_{\frac{1}{2}} = \frac{0.693}{λ} = = \frac{0.693}{0.05265} = 13.2 h$

Hence the half-life of iodine- $123$ is $13.2$ h.

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Most popular questions from this chapter

Xenon-133 is used to test lung function; it decays by emitting a beta particle.

a. Write an equation for the beta decay of Xe-133.

b. If the half-life of Xe-133 is 5.2h, how much of a 20 .-mCi sample is still active after 15.6 h?

The activity of $K - 40$ in a $70 . - kg$ human body is estimated to be $120 nCi$ . What is this activity in becquerels? $(5.3)$

Match each property ( $1 t o 3$ ) with its unit of measurement.

$1 .$ activity

$2 .$ absorbed dose

$3 .$ biological damage

a. $m r a d$

b. $g r a y$

c. $b e c q u e r e l$

d. $S v$

Identify each of the following as alpha decay, beta decay, positron emission, or gamma emission: $(5.1, 5.2)$

a. ${}_{55}^{127}{Cs} ⟶ {}_{54}^{127}{Xe} + {}_{+ 1}^{0}e$

b. ${}_{38}^{90}{Sr} ⟶ {}_{39}^{90}Y + {}_{- 1}^{0}e$

c. ${}_{85}^{218}{At} ⟶ {}_{83}^{214}{Bi} + {}_{2}^{4}{He}$

Identify each of the following as alpha decay, beta decay, positron emission, or gamma emission: $(5.1, 5.2)$

a. ${}_{13}^{27 m} Al ⟶ \frac{27}{13} Al + {}_{0}^{0}γ$

b. ${}_{5}^{8}B \to {}_{4}^{8}{Be} + {}_{- 1}^{0}e$

c. ${}_{86}^{220}{Rn} ⟶ {}_{84}^{216}{Po} + {}_{2}^{4}{Hc}$

See all solutions

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