A radioactive material has an initial amount 16 gm. After 120 days it reduces to 1 gm, then the half-life of radioactive material is
A. 60 days
B. 30 days
C. 40 days
D. 240 days

Hint: Half life of a radioactive substance is given as the time duration after which the radioactive substance decays to its half value. In this problem we have to use the Relation between the number of atoms(N) at an instant and the number of remaining atoms which is given by radioactive disintegration law.

Formula used:
Relation between number of atoms at an instant and number of remaining atoms is given as,
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\,\]
Here, N = Number of atoms at time t, \[{N_0} = \] Initial number of atoms present and \[n = \dfrac{t}{T}\,\]

Complete step by step solution:
Initial and final amount of a radioactive substance is given as 16 gm and 1 gm respectively; we have to find the half life of the substance. Using law of radioactive disintegration we get relation between number of atoms at an instant and number of remaining atoms as,
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\,.....(1)\]
Equation (1) can be rewritten as,
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}\,.....(2)\]

Given here we have,
N = 1 gm, \[{N_0} = 16\,{\rm{gm}}\] and t = 120 days
Substituting the above values in equation (2) we get,
\[\dfrac{1}{{16}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{120}}{T}}}\]
Above equation can be rewritten as,
\[{\left( {\dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{120}}{T}}}\]
As the base on both sides of the equation is same therefore power will also be same,
\[4 = \dfrac{{120}}{T}\]
$\therefore T = 30$ days
Hence, the half life of the given radioactive substance is 30 days.

Hence, option B is the correct answer.

Note: Alternatively half life of the given substance can be calculated by using equation \[N = {N_0}{e^{ - \lambda t}}\], using \[\lambda = \dfrac{{0.693}}{T}\] in the equation then take “ln” both sides of the equation.