Chapter 24, Problem 55A

Interpretation Introduction

Interpretation:

The mass of the nuclide that will remain after 49.2 y and after 98.4 y needs to be determined.

Concept introduction:

Remaining amount of radioactive element can be calculated using the formula,

$\text{N}\text{=}{\text{N}}_{\text{0}}{\left(\frac{\text{1}}{\text{2}}\right)}^n$

Where, N is the remaining amount

${\text{N}}_{\text{0}}$ is the initial amount

n is the number of half-lives that have passed.

Answer to Problem 55A

Mass of the nuclide that will remain after 49.2 y is 3 mg and after 98.4 y is 0.187 mg.

Explanation of Solution

Let the mass of nuclide remained after 49.2 y = A and the mass of nuclide remained after 98.4 y = B

Given,

Half-life of tritium $\left(\text{H}\right)$ = 12.3 y and

Mass of tritium = 48.0 mg.

Now, we divide the number of years by the value of half-life.

$\text{A:}\frac{\text{49}\text{.2}}{\text{12}\text{.3}}\text{=}\text{4 }\text{and}\text{B:}\frac{\text{98}\text{.4}}{\text{12}\text{.3}}\text{=}\text{8}$

Now, we calculate the ratio of the mass left using the following formula,

$\text{N}\text{=}{\text{N}}_{\text{0}}{\left(\frac{\text{1}}{\text{2}}\right)}^n$

Where, N is the remaining amount

${\text{N}}_{\text{0}}$ is the initial amount

n is the number of half-lives that have passed.

Therefore,

$\text{A:}{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{4}}\text{=}\frac{\text{1}}{\text{16}}\text{and}\text{B:}{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{8}}\text{=}\frac{\text{1}}{\text{256}}$

Now, we divide the mass by the ratio from the above values.

$\text{A:}\frac{\text{48}}{\text{16}}\text{=}\text{3}\text{and}\text{B:}\frac{\text{48}}{\text{256}}\text{=}\text{0}\text{.1875}$

Therefore, $\text{A=}\text{3}\text{and}\text{B}\text{=}\text{0}\text{.1875}$

Conclusion

3mg of the nuclide will remain after 49.2 y and 0.1875 mg of the nuclidewill remain after 98.4 y.