Question

The half-life of a radioactive isotope is 5.5 h. If there are initially 48 × 10³² atoms of this isotope, the number of atoms of the isotope remaining after 22 h is:
1. 3 × 10³²
2. 6 × 1032
3. 12 × 1032
4. 6 × 10⁴

Answer 1

Correct Answer - Option 1 : 3 × 10³²

Concept:

Radioactivity: It refers to the particles which are emitted from nuclei as a result of nuclear instability, it is called radioactivity.

The half-life of radioactive substance: The term half-life is defined as the time it takes for one-half of the atoms of radioactive material to disintegrate called the half-life period.

The probability of decay per unit time of radioactive material is called decay constant.

\( \lambda = \frac{{ln2}}{{{t_{\frac{1}{2}}}}} \)

Where t1/2 is half-life.

For half-life in exponential decay:

\( \frac{N}{{{N_0}}} = {e^{\lambda t}} \)

Where N = isotope remaining after decaying, NO = Original amount of isotope, λ = decay constant = 0.693/T, and T = half-life

Explanation:

Given that, t_1/2 = 5.5 h, N₀ = 48 ×10³² , t = 22 h

We have a formula,

\(\left( {\bf{\lambda }} \right) = \frac{{ln2}}{{{t_{\frac{1}{2}}}}}\)

substituting all the given values,

\(\lambda = \frac{{\ln 2}}{{{t_{1/2}}}} = \frac{{0.693}}{{5.5}} = 0.1260\;{h^{ - 1}}\)

For half-life in an exponential decay,

\( \frac{N}{{{N_0}}} = {e^{\lambda t}} \)

\( N = {N_0}{e^{\lambda t}} = 48 × {10^{32}} × {e^{ - 0.693 × \frac{{22}}{{5.5}}}} \)

\(N= 3 × {10^{32}}\)

The number of atoms of the isotope remaining after 22 h is 3 × 10³².