Q. 5.77

Question

Calcium- $47$ , used to evaluate bone metabolism, has a half-life of $4.5$ days. $(5.2, 5.4)$

a. Write the balanced nuclear equation for the beta decay of calcium- $47$ .

b. How many milligrams of a $16 - m g$ sample of calcium- $47$ remain after $18$ days?

c. How many days have passed if $4.8 mg$ of calcium- $47$ decayed to $1.2 mg$ of calcium- $47$ ?

Step-by-Step Solution

Verified

Answer

Option a is

The following is the whole nuclear reaction:

${}_{20}^{47}{Ca} \to {}_{21}^{47}{Sc} + {}_{- 1}^{0}e$

Option b is

As a result, after $18$ days, $1.0 m g C a - 47$ people remain.

Option c is

As a result, it takes $9.0$ days to decay a sample from $4.8 m g$ to $1.2 m g .$

1Step 1: Given Information (Option a)

Calcium is a periodic table $g r o u p I I$ element with the atomic symbol Ca. The alkaline earth metals contain calcium, which has an atomic number of $20$ . Calcium's beta decay results in the production of a new element with the formula ${}_{21}^{47}{Sc}$

2Step 2: Given Explanation(Option a)

The following is an incomplete nuclear equation for beta decay:

${}_{20}^{47}{Ca} \to ? + {}_{- 1}^{0}e$

Calcium has a mass number of $20$ , which is equal to the sum of the new nucleus' and a beta particle's mass numbers.

$M a s s n u m b e r o f c a l c i u m = m a s s n u m b e r o f n e w n u c l e u s + m a s s n u m b e r o f b e t a p a r t i c l e R e a r r a n g e t h e a b o v e e q u a t i o n : M a s s n u m b e r o f n e w n u c l e u s = m a s s n u m b e r o f c a l c i u m - m a s s n u m b e r o f b e t a p a r t i c l e S u b s t i t u t e t h e v a l u e s i n t h e a b o v e e q u a t i o n : M a s s n u m b e r o f n e w n u c l e u s = 47 - 0 = 47$

3Step 3: Calculus of Option a

In the equation above, find the missing atomic number:

The atomic number of calcium is $20$ , which is equivalent to the atomic number of a new nucleus plus a beta particle.

$A t o m i c n u m b e r o f c a l c i u m = (a t o m i c n u m b e r o f n e w n u c l e u s) - (a t o m i c n u m b e r o f b e t a p a r t i c l e)$

Rearrange the equation as follows:

$A t o m i c n u m b e r o f n e w n u c l e u s = (a t o m i c n u m b e r o f c a l c i u m) + (a t o m i c n u m b e r o f b e t a p a r t i c l e)$

Replace the following values in the equation:

$A t o m i c n u m b e r o f n e w n u c l e u s = 20 + 1 = 21$

4Step 4: Conclusion of Option a

Determine the new element's symbol:

Scandium $(S c)$ has atomic number $21$ in the periodic table.

The nucleus of this isotope of $(S c)$ is denoted as ${}_{21}^{47}{Sc}$

The following is the whole nuclear reaction:

${}_{20}^{47}{Ca} \to {}_{21}^{47}{Sc} + {}_{- 1}^{0}e$

5Step 5: Given Information (Option b)

(b) Half-life: A radioisotope's half-life is the time it takes for half of a sample to decay.

The half-life of calcium- $47$ is $4.5$ days.

The starting dose of calcium- $47$ is $16 m g .$

Make a plan to figure out the unknown quantity.

days $C a - 47$ half-life number of half-lives miligrams Number of remaining half-lives in miligrams of $C a - 47$

6Step 6: Given Explanation (Option b)

Calculate the amount of sample left after $18$ days using the using conversion factor. Calcium's half-life is $47 = 4.5$ days.

$\frac{h a l f - l i f e o f c a l c i u m - 47}{4.5 d a y s} a n d \frac{4.5 d a y s}{h a l f - l i f e o f c a l c i u m - 47}$

To begin, calculate the number of half-lives in the period that has passed.

$Number of half - lives = 18 days \times \frac{1 half - life of calcium - 47}{4.5 days} = 4 half - lives of Ca - 47$

Calculate how much of the sample decays in four half-lives and how much calcium remains in milligrammes.

$16 mg Ca - 47 \overset{1 half - life}{\to} 8.0 mg Ca - 47 \overset{2 half - lives}{\to} 4.0 mg Ca - 47 \overset{3 half - lives}{\to} 2.0 mg Ca - 47 \overset{4 halt - lives}{\to} 1.0 mg Ca - 47$