Question

Certain radio-active substance reduces to 25% of its value in 16 days. Its half-life is
A. 32 days
B. 8 days
C. 64 days
D. 28 days

Answer 1

Hint: Time interval in which a radioactive element decays to its half is called it half life, i.e. for \[N = \dfrac{{{N_0}}}{2}\], time \[t = {t_{1/2}}\] and \[N = {N_0}{e^{ - \lambda T}}\] where, T is half life and \[\lambda \] is decay constant.

Formula used:
\[N = {N_0}{e^{ - \lambda T}}\] and \[\lambda = \dfrac{{\ln 2}}{T}\]
Here, N = Number of atoms after time t, \[{N_0} = \] Initial number of atoms, \[\lambda = \] Decay Constant and T = Half life

Complete step by step solution:
Given here is a radioactive substance which reduces to 25% of its value, we have to find the half life of the substance. The initial number of atoms in the given radioactive substance is \[{N_0}\]then the remaining number of atoms N in the substance will be,
N = 25% of or \[\dfrac{{25}}{{100}} \times {N_0} = \dfrac{{{N_0}}}{4}\,........(1)\]
As we know that, \[N = {N_0}{e^{ - \lambda T}}......(2)\]

Substituting, \[N = \dfrac{{{N_0}}}{4}\,\]from equation (1) and t = 16 days in equation (2) we get,
\[\dfrac{{{N_0}}}{4} = {N_0}{e^{ - 16\lambda }} \\
\Rightarrow \dfrac{1}{4} = {e^{ - 16\lambda }}\]
Above equation can be written as,
\[\ln 4 = 16\lambda \\
\Rightarrow \lambda = \dfrac{{\ln 4}}{{16}}\,......(3)\]

As we know that decay constant is given by,
\[\lambda = \dfrac{{\ln 2}}{T}\]
Equating \[\lambda = \dfrac{{\ln 2}}{T}\] with equation (4) we get,
\[\dfrac{{\ln 2}}{T} = \dfrac{{\ln 4}}{{16}}\,\]
This can be further simplified as,
\[\dfrac{{\ln 2}}{T} = \dfrac{{2\ln 2}}{{16}} \\
\therefore T = \dfrac{{16}}{2} = 8\,{\rm{days}}\]
Hence, the half life of the given substance is 8 days.

Henced, option B is the correct option.

Note: Decay constant is defined as the probability of decay for radioactive nuclei per unit time and it is independent of temperature, pressure and strength of the bonds by which a radioactive element is held. Decay constant decreases with time.